1. (20 points) Finding critical points, maximums, minimums.

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1 Math 030 (Section 02) Test #3 June 2 st, 200 Name: ANSWER KEY Be sure to show your work!. (20 points) Finding critical points, maximums, minimums. (a) Let f(x) = 2x 3 + 5x 2 4x 3. Find min/max on [, 2] and all critical points. First, let s plot f(x) to know what we re dealing with. Alpha: plot 2x^3+5x^2-4x-3 Alpha: plot 2x^3+5x^2-4x-3 on [-,2] To find the critical points of f(x) we need to compute its derivative. Alpha: derivative 2x^3+5x^2-4x-3 The derivative of f(x) has roots (i.e. f (x) = 0) at x = 2 and x = /3. Since f (x) is defined everywhere (the formula doesn t have any divisions, logs, square roots, etc. to worry about), we don t have any additional critical points coming from an undefined derivative. The critical points of f(x) are located where x = 2 and /3. To find the maximum of f(x) on the interval [, 2], we need to check the value of f(x) at all of its critical points lying between and 2 (so /3 but not 2) as well as the endpoints and 2 themselves. In Alpha we can plug in values one at a time using... at x=- or... where x=- or we can plug in all of the values at once as follows: Alpha: 2x^3+5x^2-4x-3 at x={-,/3,2} and click on Approximate form. We get the following values: x = - /3 2 f(x) = The maximum value of f(x) on the interval [, 2] is 25. The maximum is located at x = 2. The minimum value of f(x) on the interval [, 2] is The minimum is located at x = /3 = e x2 x+ x 0 (b) Let g(x) = 2x 3 + 5x 2 4x 3 x > 0 Again, let s plot our function to see what we ve got. Alpha: Piecewise[{ {e^(-x^2-x+),x<=0}, {2x^3+5x^2-4x-3,x>0} }] Alpha: Piecewise[{ {e^(-x^2-x+),x<=0}, {2x^3+5x^2-4x-3,x>0} }] on [-3,]

2 Note: Alpha is very sensitive about long complicated inputs especially those involving Piecewise. Don t count on Alpha always doing what you want when you ask it to preform some task on a piecewise defined function. To find the critical points of g(x), we need to examine its derivative. Let s examine each piece of this piecewise function one at a time. Alpha: derivative e^(-x^2-x+) The derivative is defined everywhere and has a single root at x = /2 (which is within this formula s bound of x 0). Next, the function is discontinuous where we switch formulas, so we get a critical point there (at x = 0). Then recall that the second formula had critical points at and /3. x = /3 is within this formula s bound of x > 0 but x = is not (so we throw it out). The critical points of g(x) are located where x = /2, 0, and /3. To find the maximum of g(x) on the interval [ 3, ], we need to check the value of g(x) at all of its critical points lying between 3 and (that is all of them) as well as the endpoints 3 and themselves. Since our function is piecewise defined, we need to pay careful attention to which values we plug into which formulas. x = 3, /2, and 0 should be plugged into e x2 x+ since it s valid for all x 0. x = /3 and should be plugged into 2x 3 + 5x 2 4x 3 because it s valid for x > 0. Alpha: e^(-x^2-x+) at x={-3,-/2,0} and click on Approximate form. Alpha: 2x^3+5x^2-4x-3 at x={/3,} and click on Approximate form. We get the following values: x = -3 -/2 0 /3 g(x) = The maximum value of g(x) on the interval [ 3, ] is The maximum is located at x = /2 = 0.5. The minimum value of g(x) on the interval [ 3, ] is The minimum is located at x = /3 = Technical Note: When encountering a discontinuity, we should really check right and left hand limits as well as the function s value at the discontinuity. In our problem this does not effect the answer, but it s not hard to come up with examples where it does. Looking at a good graph can keep us from worrying about this too much. Our answers match what we see on the graph, so we won t worry about weirdness at the discontinuity. 2. (8 points) Given the following data, compute elasticity (elasticity for columns 2 and 3 should be located in column 3). Circle E if the situation is elastic, I if inelastic, and U if unitary. Since we can computing multiple elasticities, Excel is the best choice of software. Punch in the values in the table and then (assuming the q = is in cell A, is in cell B, p = is in cell A2, $30 is in cell B2, etc.) type (in cell B3) the following formula: = -( (C-B)*(B2+C2) ) / ( (C2-B2)*(B+C) ) and then copy it across. Be careful to get your parentheses matched up correctly!! Once we compute the elasticities, remember that E < is inelastic, E > is elastic, and E = is unitary. q = p = $30 $5 $7.50 $4 $2 Elasticity = Elastic Unitary Inelastic Inelastic 2

3 3. (4 points) Point Elasticity (a) Kyle s owns a rental store. He has found that his demand function for a daily rental of his small dump truck is p = 5 ln(q) If Kyle sets his price at $ To find the number of customers (quantity unless somebody tries to rent more than one dump truck...hmm) we need to solve the equation 300 = 5ln(q) for q. Alpha: 300 = -5ln(q) and click on Approximate form. We get that p(q) = $300 when q Kyle will have approximately 25 customers. To find the point elasticity when q = and p = $300 we need to differentiate the demand function, plug in q and then put it all together. Alpha: derivative -5ln(q) at q= Alpha: -(300/24.963)/( ) His point elasticity would be ε = Since ε = >, the situation is elastic (quantities dominate not prices), so revenue and price move in opposite directions. If Kyle raises his price, his revenue will decrease. (b) Let ε =.25 (point elasticity) for some quantity and price data. If this price is decreased by 5%... Remember that elasticity is negative percent change in quantity divided by percent change in price. We percent change in quantity are told that the percent change in price is 5% so that.25 =. Therefore, 0.05 the percent change in quantity is.25(0.05) = which is 6.25%....the corresponding quantity will increase by 6.25%. 4. (6 points) George has opened a coffee stand supplying a quick cup of coffee to all that pass by. George can obtain decent coffee for $8 a pound, but pays inventory cost of $.50 per pound per year (base inventory costs on the average amount in inventory with all of the standard assumptions). In addition suppose George must pay $25 to get his coffee delivered and he will need,000 lbs. of coffee each year. (a) Find George s economic order quantity (EOQ). ( x ) ( ) 000 George s annual cost function is C(x) = 000(8) where x is size of his orders. 2 x Let s graph C(x) to see what we re dealing with. Alpha: plot 000(8) +.50(x/2) + 25(000/x) on [0,000] It looks like our minimum occurs somewhere around 200. To find the exact answer we need to locate the critical points of C(x). 3

4 Alpha: derivative 000(8) +.50(x/2) + 25(000/x) Our function (and its derivative) is undefined at x = 0. Also, C (x) = 0 when x = ± (the roots of the derivative). Only x = is within our range of values we re considering and it s close to our guess for the location of the minimum, so we have that the EOQ is x = To find the corresponding minimum annual cost... Alpha: 000(8) +.50(x/2) + 25(000/x) at x= George s ideal EOQ is lbs. of coffee. His ideal minimum annual cost is $8, (b) Now consider what happens if George gets a discount when making a large order. Suppose that when George makes an order of 300 lbs. or more, he get the coffee for $7 a pound, but must pay $50 to place an order. His inventory costs stay the same. In this case... ) ( ) 000(8) +.50 ( x 2 George s annual cost function is C(x) = 000(7) +.50 ( x 2 Let s graph C(x) to see what we re dealing with. x ) + 50 ( 000 x ) x < 300 x 300 Alpha: Piecewise[{{000(8)+.50(x/2)+25(000/x),x<300}, {000(7)+.50(x/2)+50(000/x),x>=300}}] on [0,000] Alpha: plot 000(7)+.50(x/2)+50(000/x) on [300,000] The graph on the left clearly shows that the minimum occurs when we make large orders of 300 or more lbs. So we can discard the first formula and just deal with the second. The graph on the right indicates that the minimum occurs when x = 300. Let s find the critical points of this function to confirm our suspicion. Alpha: derivative 000(7)+.50(x/2)+50(000/x) We find that this derivative is 0 when x = ± both of which are out of bounds (less than 300), so our minimum does indeed occur at x = 300. We need to plug this into C(x) to find the corresponding minimal cost. Of course since x = 300 we use the second formula. Alpha: 000(7)+.50(x/2)+50(000/x) at x=300 George s ideal EOQ is 300. His ideal minimum annual cost is $7, Note: Although we could order 300 lbs. of coffee at a time, this solution is still not practical since we would need to make 000/300 = orders per year (unless we allow inventory to spill over from one year to the next). 5. (4 points) Consider the function f(x) = e x2. (a) Using the right hand rule and 5 rectangles, approximate the area under the graph of y = e x2 2 x 3. when Open up the area_approx.xls right hand rule template in Excel. We are given a = 2, b = 3, and n = 5. Enter the formula = EXP(-(B4^2)) in cell C4 and copy down. Don t forget the parentheses to take care of the Excel sign error. Area

5 (b) Using the graph below sketch what the left hand rule approximation with n = 2 looks like. Created in Maple using: with(student[calculus]); ApproximateInt(exp(-x^2), x = 2.. 3, method = left, partition = 2, output = plot); In general, the left hand rule approximation will give an overestimate. function is decreasing between 2 and 3. This happens because the To find the exact are we need to use an integral. Alpha: int e^(-x^2) from x=2 to 3 (c) The exact area under y = e x2 when 2 x 3 is (round to 3 decimals). 6. (3 points) Were-wolfe (a very talented duet) released their first album on January, The album s sales rate can be modeled by the function S(t) = 5e t (where t is the number years since the album s release and S(t) is the sales rate in millions of albums per year). S(t) is a sales-rate function, so to find how many albums will be sold we need to integrate it. [The units of S(t) are millions of albums per year, so the units of it integrals will be millions of albums.] To figure out what interval to integrate over, notice that 2008 corresponds to 0 t, 2009 is t 2 and so this year, 200, is 2 t 3. We need to compute Alpha: int 5e^(-sqrt(t)) from t=2 to 3 This evaluates to (millions of albums). (a) Were-wolfe will sell,035,780 albums this year (200). 3 2 S(t) dt. 5

6 Our next task is to find out how long it takes to sell a certain number of albums. Since we are asking when? we will need to solve for a limit of an integral. In fact, we need to solve the following equation: T 0 S(t) dt = 8 then T will be the number of years it takes to sell 8 million albums. Alpha: int 5e^(-sqrt(t)) from 0 to T click on the solution and copy the formula appearing after the equals sign. Alpha: 5 (2-2 e^(-sqrt(t)) (sqrt(t)+)) = 8 We get a solution of T So in the eighth year the album is out the 8 million-th copy will sell = 206 Note: Don t round. For example: if it takes 0.9 years to sell a certain number of copies, then this is done within the first year, 2008 not = (b) Were-wolfe will sell their 8 million-th album in the year (5 points) Everyone in the Lambert family has a pet dog. The dogs weights are distributed normally with a mean of 75 lbs. and standard deviation of 0 lbs. There are a total of 200 dogs owned by the extended Lambert family. (a) How many dogs weigh between 50 and 60 lbs.? First, we need to compute the probability that a dog weighs between 50 and 60 lbs. This is computed by the following integral: P (50 X 60) = π(02 ) e (x 75)2 /(2(0)) dx Alpha: normal distribution click on the formula for the normal distribution and copy it. change mu to 75 and sigma to 0. Alpha: int e^(-(x-75)^2/(2 0^2))/(sqrt(2 pi) 0) from x=50 to 60 Then So we have that P (50 X 60) So about 6.06/This means that = 2.95 dogs have that weight. Answer: 2 dogs weigh between 50 and 60 lbs. (b) How many dogs weigh under 20 lbs.? First, we need to compute the probability that a dog weighs under 20 lbs. This is computed by the following integral: P (X 20) = 20 2π(02 ) e (x 75)2 /(2(0)) dx Alpha: int e^(-(x-75)^2/(2 0^2))/(sqrt(2 pi) 0) from x=-infinity to 20 So we have that P (X 20) This means that ( ) dogs have that weight. Answer: No dog weighs less than 20 lbs. (c) What weight is the cut-off for the heaviest 0% of the Lambert s dogs? We want to find a weight W such that P (W X) = /(2(0)) dx = 0.0. First we W 2π(02 ) e (x 75)2 will compute the integral and then solve the corresponding equation. Alpha: int e^(-(x-75)^2/(2 0^2))/(sqrt(2 pi) 0) from x=w to infinity click on the result and copy the formula to the right of the equality. Alpha: /2 erfc((w-75)/(0 sqrt(2))) = 0.0 Answer: The heaviest 0% of the dogs weigh lbs. or more. 6

7 Math 030 (Section 03) Test #3 June 2 st, 200 Name: ANSWER KEY Be sure to show your work!. (20 points) Finding critical points, maximums, minimums. (a) Let f(x) = 3x 3 4x 2 7x + 6. Find min/max on [ 2, 4] and all critical points. First, let s plot f(x) to know what we re dealing with. Alpha: plot 3x^3-4x^2-7x+6 Alpha: plot 3x^3-4x^2-7x+6 on [-2,4] To find the critical points of f(x) we need to compute its derivative. Alpha: derivative 3x^3-4x^2-7x+6 The derivative of f(x) has roots (i.e. f (x) = 0) at x = and x = 7/9. Since f (x) is defined everywhere (the formula doesn t have any divisions, logs, square roots, etc. to worry about), we don t have any additional critical points coming from an undefined derivative. The critical points of f(x) are located where x = and 7/ To find the maximum of f(x) on the interval [ 2, 4], we need to check the value of f(x) at all of its critical points lying between 2 and 4 (that is all of them) as well as the endpoints 2 and 4 themselves. In Alpha we can plug in values one at a time using... at x=-2 or... where x=-2 or we can plug in all of the values at once as follows: Alpha: 3x^3-4x^2-7x+6 at x={-2,-,7/9,4} and click on Approximate form. We get the following values: x = -2-7/9 4 f(x) = The maximum value of f(x) on the interval [ 2, 4] is 66. The maximum is located at x = 4. The minimum value of f(x) on the interval [ 2, 4] is The minimum is located at x = 7/9 = x 3 4x 2 7x + 6 x < 0 (b) Let g(x) = (x + ) 2 e x2 +x+ x 0 Again, let s plot our function to see what we ve got. Alpha: Piecewise[{{3x^3-4x^2-7x+6,x<0}, {(x+)^2e^(-x^2+x+),x>=0}}] Alpha: Piecewise[{{3x^3-4x^2-7x+6,x<0}, {(x+)^2e^(-x^2+x+),x>=0}}] on [-2,4] 7

8 Note: Alpha is very sensitive about long complicated inputs especially those involving Piecewise. Don t count on Alpha always doing what you want when you ask it to preform some task on a piecewise defined function. To find the critical points of g(x), we need to examine its derivative. Let s examine each piece of this piecewise function one at a time. Alpha: derivative (x+)^2e^(-x^2+x+) The derivative is defined everywhere and has roots at x = 3/2,, and. Only x = is within this formula s bound (x 0), so we keep it and discard the others. Next, the function is discontinuous where we switch formulas, so we get a critical point there (at x = 0). Then recall that the first formula has critical points at and 7/9. x = is within this formula s bound of x < 0 but x = 7/9 is not (so we throw it out). The critical points of g(x) are located where x =, 0, and. To find the maximum of g(x) on the interval [.5,.5], we need to check the value of g(x) at all of its critical points lying between.5 and.5 (that is all of them) as well as the endpoints.5 and.5 themselves. Since our function is piecewise defined, we need to pay careful attention to which values we plug into which formulas. x =.5 and should be plugged into 3x 3 4x 2 7x + 6 since it s valid for all x < 0. x = 0, and.5 should be plugged into (x + ) 2 e x2 +x+ because it s valid for x 0. Alpha: 3x^3-4x^2-7x+6 at x={-.5,-} Alpha: (x+)^2e^(-x^2+x+) at x={0,,.5} and click on Approximate form. We get the following values: x = g(x) = The maximum value of g(x) on the interval [.5,.5] is 6. The maximum is located at x =. The minimum value of g(x) on the interval [.5,.5] is The minimum is located at x = 0. Technical Note: When encountering a discontinuity, we should really check right and left hand limits as well as the function s value at the discontinuity. In our problem this does not effect the answer, but it s not hard to come up with examples where it does. Looking at a good graph can keep us from worrying about this too much. Our answers match what we see on the graph, so we won t worry about weirdness at the discontinuity. 2. (8 points) Given the following data, compute elasticity (leave the first box next to elasticity blank so that elasticity for columns 2 and 3 is located in column 3). Circle E if the situation is elastic, I if inelastic, and U if unitary. Since we can computing multiple elasticities, Excel is the best choice of software. Punch in the values in the table and then (assuming the q = is in cell A, is in cell B, p = is in cell A2, $30 is in cell B2, etc.) type (in cell B3) the following formula: = -( (C-B)*(B2+C2) ) / ( (C2-B2)*(B+C) ) and then copy it across. Be careful to get your parentheses matched up correctly!! Once we compute the elasticities, remember that E < is inelastic, E > is elastic, and E = is unitary. q = p = $500 $300 $75 $00 $50 Elasticity = Elastic Elastic Unitary Inelastic 8

9 3. (4 points) Point Elasticity (a) Kyle s owns a rental store. He has found that his demand function for a daily rental of his small dump truck is p = 245 ln(q) If Kyle sets his price at $ To find the number of customers (quantity unless somebody tries to rent more than one dump truck...hmm) we need to solve the equation 350 = 245ln(q) for q. Alpha: 350 = -245ln(q) and click on Approximate form. We get that p(q) = $350 when q Kyle will have approximately 64 customers. To find the point elasticity when q = and p = $350 we need to differentiate the demand function, plug in q and then put it all together. Alpha: derivative -245ln(q) at q=64.28 Alpha: -(350/64.28)/(-3.839) His point elasticity would be ε =.429. Since ε =.429 >, the situation is elastic (quantities dominate not prices), so revenue and price move in opposite directions. If Kyle raises his price, his revenue will decrease. (b) Let ε = 0.75 (point elasticity) for some quantity and price data. If this price is increased by 3%... Remember that elasticity is negative percent change in quantity divided by percent change in price. We percent change in quantity are told that the percent change in price is 3% so that 0.75 =. Therefore, 0.03 the percent change in quantity is 0.75( 0.03) = which is 2.25%....the corresponding quantity will decrease by 2.25%. 4. (6 points) George has opened a coffee stand supplying a quick cup of coffee to all that pass by. George can obtain decent coffee for $2 a pound, but pays inventory cost of $2 per pound per year (base inventory costs on the average amount in inventory with all of the standard assumptions). In addition suppose George must pay $50 to get his coffee delivered and he will need 5,000 lbs. of coffee each year. (a) Find George s economic order quantity (EOQ). ( x ) ( ) 5000 George s annual cost function is C(x) = 5000(2) where x is size of his orders. 2 x Let s graph C(x) to see what we re dealing with. Alpha: plot 5000(2) + 2(x/2) + 50(5000/x) on [0,5000] It looks like our minimum occurs somewhere around 500. To find the exact answer we need to locate the critical points of C(x). 9

10 Alpha: derivative 5000(2) + 2(x/2) + 50(5000/x) Our function (and its derivative) is undefined at x = 0. Also, C (x) = 0 when x = ±500 (the roots of the derivative). Only x = 500 is within our range of values we re considering and it matches our guess for the location of the minimum, so we have that the EOQ is x = 500. To find the corresponding minimum annual cost... Alpha: 5000(2) + 2(x/2) + 50(5000/x) at x=500 George s ideal EOQ is 500 lbs. of coffee. His ideal minimum annual cost is $6,000. Note: This solution is not just ideal, it s also practical. George should order 500 lbs of coffee 0 times each year. (b) Now consider what happens if George gets a discount when making a large order. Suppose that when George makes an order of 000 lbs. or more, he get the coffee for $9 a pound, but must pay $00 to place an order. His inventory costs stay the same. In this case... ) ( ) 5000(2) + 2 ( x 2 George s annual cost function is C(x) = 5000(9) + 2 ( x 2 Let s graph C(x) to see what we re dealing with. x ) + 00 ( 5000 x ) x < 000 x 000 Alpha: Piecewise[{{5000(2)+2(x/2)+50(5000/x),x<000}, {5000(9)+2(x/2)+00(5000/x),x>=000}}] Alpha: plot 5000(9)+2(x/2)+00(5000/x) on [000,5000] The graph on the left clearly shows that the minimum occurs when we make large orders of 000 or more lbs. So we can discard the first formula and just deal with the second. The graph on the right indicates that the minimum occurs when x = 000. Let s find the critical points of this function to confirm our suspicion. Alpha: derivative 5000(9)+2(x/2)+00(5000/x) We find that this derivative is 0 when x = ± both of which are out of bounds (less than 000), so our minimum does indeed occur at x = 000. We need to plug this into C(x) to find the corresponding minimal cost. Of course since x = 000 we use the second formula. Alpha: 5000(9)+2(x/2)+00(5000/x) at x=000 George s ideal EOQ is 000. His ideal minimum annual cost is $46,500. Note: This solution is not just ideal, it s also practical. George should order 000 lbs of coffee 5 times each year. 5. (4 points) Consider the function f(x) = e x2. (a) Using the right hand rule and 0 rectangles, approximate the area under the graph of y = e x2 0 x 0.5. when Open up the area_approx.xls right hand rule template in Excel. We are given a = 0, b = 0.5, and n = 0. Enter the formula = EXP(-(B4^2)) in cell C4 and copy down. Don t forget the parentheses to take care of the Excel sign error. Area

11 (b) Using the graph below sketch what the trapezoid rule approximation with n = 2 looks like. Created in Maple using: with(student[calculus]); ApproximateInt(exp(-x^2), x = , method = trapezoid, partition = 2, output = plot); In general, the trapezoid hand rule approximation will give an underestimate. This happens because the function is concave down between 0 and 0.5. To find the exact are we need to use an integral. Alpha: int e^(-x^2) from x=0 to 0.5 (c) The exact area under y = e x2 when 0 x 0.5 is 0.46 (round to 3 decimals). 6. (3 points) Were-wolfe (a very talented duet) will release their first album on January, 20. The album s sales rate will be modeled by the function S(t) = te t (where t is the number years since the album s release and S(t) is the sales rate in millions of albums per year). S(t) is a sales-rate function, so to find how many albums will be sold we need to integrate it. [The units of S(t) are millions of albums per year, so the units of it integrals will be millions of albums.] To figure out what interval to integrate over, notice that 20 corresponds to 0 t, 202 is t 2, etc. So 204 is 3 t 4. We need to compute 4 Alpha: int te^(-sqrt(t)) from t=3 to 4 This evaluates to (millions of albums). 3 S(t) dt. (a) Were-wolfe will sell 538,02 albums during the year 204.

12 Our next task is to find out how long it takes to sell a certain number of albums. Since we are asking when? we will need to solve for a limit of an integral. In fact, we need to solve the following equation: T 0 S(t) dt = 5 then T will be the number of years it takes to sell 5 million albums. Alpha: int te^(-sqrt(t)) from 0 to T click on the solution and copy the formula appearing after the equals sign. Alpha: 2-2 e^(-sqrt(t)) (T^(3/2)+3 T+6 sqrt(t)+6) = 5 We get a solution of T So in the tenth year the album is out the 5 million-th copy will sell = 202 Note: Don t round. For example: if it takes 0.8 years to sell a certain number of copies, then this is done within the first year, 20 not 20 + = 202. (b) Were-wolfe will sell their 5 million-th album in the year (5 points) Everyone in the Lambert family has a pet dog. The dogs weights are distributed normally with a mean of 50 lbs. and standard deviation of 5 lbs. There are a total of 300 dogs owned by the extended Lambert family. (a) How many dogs weigh between 80 and 90 lbs.? First, we need to compute the probability that a dog weighs between 50 and 60 lbs. This is computed by the following integral: P (80 X 90) = π(52 ) e (x 50)2 /(2(5)) dx Alpha: normal distribution click on the formula for the normal distribution and copy it. change mu to 50 and sigma to 5. Alpha: int e^(-(x-50)^2/(2 5^2))/(sqrt(2 pi) 5) from x=80 to 90 Then So we have that P (80 X 90) So about.89/this means that = dogs have that weight. Answer: 6 dogs weigh between 80 and 90 lbs. (b) How many dogs weigh under 0 lbs.? First, we need to compute the probability that a dog weighs under 0 lbs. This is computed by the following integral: P (X 0) = 0 2π(52 ) e (x 50)2 /(2(5)) dx Alpha: int e^(-(x-50)^2/(2 5^2))/(sqrt(2 pi) 5) from x=-infinity to 0 So we have that P (X 0) This means that =.494 dogs have that weight. Answer: Only dog weighs less than 0 lbs. (c) What weight is the cut-off for the heaviest 5% of the Lambert s dogs? We want to find a weight W such that P (W X) = /(2(5)) dx = First we W 2π(52 ) e (x 50)2 will compute the integral and then solve the corresponding equation. Alpha: int e^(-(x-50)^2/(2 5^2))/(sqrt(2 pi) 5) from x=w to infinity click on the result and copy the formula to the right of the equality. Alpha: /2 erfc((w-50)/(5 sqrt(2))) = 0.05 Answer: The heaviest 5% of the dogs weigh lbs. or more. 2