Math 120, Winter Answers to Unit Test 3 Review Problems Set B.

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1 Math 0, Winter 009. Answers to Unit Test Review Problems Set B. Brief Answers. (These answers are provided to give you something to check your answers against. Remember than on an eam, you will have to provide evidence to support your answers and you will have to eplain your reasoning when you are asked to.). There are many possible answers to this problem. What we are looking for here is that you have the right features (y-intercept, increasing/decreasing and concavity) in the right places. The graph of one function that has the required features is shown below.. If the hotel chain takes only 00 chairs then the total revenue will be $7,000. If the hotel chain takes chairs (where is between 00 and 400) then the revenue is given by: R() = ( ( 00)). This problem illustrates the phenomenon that coming up with the function that describes the situation is often the hardest part of an optimization problem. The rationale behind this complicated formula are as follows: First, revenue will be equal to the number of chairs times the price of each chair. Therefore, R() = (Price). The basic price of each chair is $90. However, this price is lowered by twenty-five cents for every chair (above 00 chairs) purchased. If chairs are purchased, but the first 00 do not create a discount, then the number of chairs that create discounts will be: 00. Each of these chairs creates a twenty-five cent discount, so in units of dollars, the total discount on the chair price will be: 0.5 ( 00). Subtracting this discount from the $90 basic chair price gives the discounted chair price of: ( 00). Multiplying this price by, the number of chairs actually sold, gives the revenue, R(). Differentiating this function and setting the derivative equal to zero gives = 0. To see this, observe that by FOILing, you can simplify the formula for R().

2 ( ) = " 90 # 0.5" ( # 00) R ( ) =65 " # 0.5". R "( ) =65 # 0.5 $ Checking the sign of the derivative on each side of = 0 confirms that this is a local maimum. X Derivative When = 0, the revenue will be R(0) = $7,5. The one other point to check is the endpoint = 400. The revenue when = 400 is: R(400) = $6,000. Therefore, the greatest revenue that the furniture company can collect is $7,5 (when 0 chairs are sold) and the lowest is $6,000 (when 400 chairs are sold). Astonishingly, the company collects more revenue when they sell 00 chairs than when they sell 400 chairs. As the cost of procing 400 chairs will almost certainly be significantly higher than the cost of procing 00 chairs, the company will make much smaller profits from selling 400 chairs, as compared with their profits from selling 00 chairs. That is not very good business practice, so they might want to re-think their pricing strategy..(a) The sum # f ( k " ) " represents the approimation to the value of the true value of the integral: k= 0 8 " f ( ) # d 0 that you would get if you used 4 rectangles (each with a width Δ = ) to approimate the area under the curve y = f() between = 0 and = 8..(b) The area that would have the same numerical value as the sum # f ( k " ) " is shaded in the diagram shown below. k= 0 y = f()

3 .(c) The function values needed to approimate the sum # f ( k " ) " are given in the table below. k k f ( " k) The numerical value of the sum is: k= 0 # f ( k " ) " = k= 0 ( ) " = (d) Using 00 rectangles to approimate the area between the curve y = f() and the -ais from = 0 to = 8, the relevant commands for a TI-8 calculator are: (8-0)/00 [STO ] W Y = (-/7)*(X-)*(X-.5)*(X-6) + 5 sum(seq(y(0+k*w)*w,k,0,99)) The numerical value of this approimation of the area under the curve is: (e) The symbolic epression that will represent the precise value of the area beneath the graph y = f() between = 0 and = 8 is: Precise Value of Area = 8 " f ( ) # d. 0.(f) The equation for the function f() is: f ( ) = " # 7 ( ) " ( #.5) " ( # 6) + 5 = # " 7 + " # 6 " An anti-derivative, F(), of f() is given by: F( ) = " # # " 6 # # + C. Using this anti-derivative the precise value of the area will be given by: Precise Value of Area = 8 " f ( ) # d = F(8) F(0)

4 4.(a) The critical points of F() are the places where f() = 0. These occur at = and = 5. To classify these: = : = 5: The graph of f() is negative just before = and positive just after =, so by the First Derivative test, = is a local minimum. The graph of f() is positive just to the left and just to the right of =5. This point is neither a maimum nor a minimum. 4.(b) The global minimum is located at =. The global maimum occurs at =0 and =6. 4.(c) Inflection points are where the graph of F() changes concavity. On Figure 4, this is indicated by the graph changing from increasing to decreasing or vice versa. Looking at Figure 4, you can see that this happens at =4 and =5. So, F() has points of inflection at =4 and =5. 4.(d) The graph of y = F() with F()=4 is shown below. 5.(a) In order for F() to be an anti-derivative of f(), the derivative of F() should be equal to f(). That is: Differentiating the given F case F() is not an anti-derivative of f(). F "( ) = f ( ). 5.(b) Differentiating the given function F derivatives: ( ) = yields: F "( ) = 0. This is not equal to f ( ) =, so in this ( ) = " ln( ) # using the Proct Rule for F "( ) =# ln( ) $ # $ = ln( ) $+= ln( ).

5 This is equal to f ( ) = ln ( ), so in this case F() is an anti-derivative for f(). Using this antiderivative to evaluate the definite integral 5.(c) derivatives: 0 # f ( ) " d = F 0 0 # f ( ) " d : [ ] %.64. ( ) $ F( ) =0 " ln( 0) $0 $ " ln( ) $ Differentiating the given function F This is equal to f ( ) = 0# + F "( ) = " ( + so in this case F() is an anti-derivative of f(). Using this anti- ) derivative to evaluate the definite integral 5.(d) derivatives: 0 ( ) = ( ) $ $ = ( + ) ( + ). using the Quotient Rule for + # f ( ) " d = F 0 0 # f ( ) " d : ( ) $ F( ) = Differentiating the given function F F "( ) = +0 $ + % $0.9 ( ) = ln ln( ) [ ln( ) ] # = # ln ( ) using the Chain Rule for ( ). This is not equal to f ( ) = ln( ) so in this case F() is not an anti-derivative of f(). 5.(e) gives: Differentiating the given function F ( ) = 0 # ln( ) $# ln( ) F " ( ) = ln( ) $ = [ ] [ ln( ) ] = $ # [ ln( ) ]. with the Quotient Rule for derivatives This is equal to f ( ) = " [ ( )] # ln so in this case F() is equal to f(). Using this antiderivative to evaluate the definite integral 0 # f ( ) " d :

6 0 # f ( ) " d = F 0 ( ) $ F( ) = ln(0) $ ln() % $ (a) You should use 7 rectangles. 6.(b) Left-hand Riemann sum = Right-hand Riemann sum = (c) The best estimate is the average of left and right sums, which is equal to (d) Midpoint sum = This is different to the answer from Part (c) of this problem. 7.(a) The table showing the work involved in carrying out Euler s method is given below. According to the figures given in this table, f() Current Current f() Derivative Rise New f() (b) The approimation f().9906 is an underestimate of the true value of f(). This is because the increasing derivatives in the table from Part (a) show us that the function f() is concave up and Euler s method gives an underestimate when the function approimated is concave up. 8.(a) The values of the function F() are shown in the table given below. 4 F() 8 π / 4 7 / π / 4 8 π / 4 7 π / 4 9.(a) # 7 + " d you can: I) Set u = 7 +. d = ln ( ) ". III) Rearrange to make d the subject of the equation: d = ln( ) ". IV) Substitute into the indefinite integral using the epressions for u and d: # " d = # " 7+ u = ln # ln( ) " ". u ( )"

7 # ln( ) " " u = ln( ) " ln( u) + C. VI) Convert the anti-derivative back to variable : # " d = 7+ ln( ) " ln( 7 + ) + C. ( ) # d 9.(b) $ ( 4 " 6) # " + 8 you can: I) Set u = " + 8. = ". d III) Rearrange to make d the subject of the equation: d = ". IV) Substitute into the indefinite integral using the epressions for u and d: $ ( 4 " 6) # ( " + 8) # d = ( 4 " 6) # u $ # " = $ u #. # u " = " u + C. VI) Convert the anti-derivative back to variable : $ ( 4 " 6) # " + 8. ( ) # d = # ( " + 8) + C 9.(c) # + you can: ( ) " d I) Set u =+. d =. III) Rearrange to make d the subject of the equation: d =. IV) Substitute into the indefinite integral using the epressions for u and d: # " d = " # ( + ) u = # " = u # " u$ ". $ " u# " = " u# # + C = # u + C. VI) Convert the anti-derivative back to variable : $ # " d = ( + ) " +. ( ) + C 9.(d) 9 $ " # d you can: I) Set u =". d = ".

8 III) Rearrange to make d the subject of the equation: d = ". IV) Substitute into the indefinite integral using the epressions for u and d: 9 $ # d = $ 9 # = "# " u " u # = "# u" $ $ #. "# u " $ # = "# u + C = "6 # u + C. 9 VI) Convert the anti-derivative back to variable : $ # d = "6 # " + C. " 9.(e) 4 # " d you can: I) Set u = d =4. III) Rearrange to make d the subject of the equation: d = 4. IV) Substitute into the indefinite integral using the epressions for u and d: 4 4 # = 7 +7 # = u 4 # " u. # " = ln u u ( ) + C. 4 VI) Convert the anti-derivative back to variable : # " d = ln ( ) + C. 9.(f) e " d # you can: I) Set u =. d = ". III) Rearrange to make d the subject of the equation: d = " ". IV) Substitute into the indefinite integral using the epressions for u and d: e " d e # = u # " " " = " e u " #. # " e u " = " e u + C. e VI) Convert the anti-derivative back to variable : # " d = " e + C. 0. The most cost-effective solution will probably be the one that involves the shortest possible pipelines. Therefore, the quantity that you should try to minimize here is the total length of the pipeline. If the variable represents the distance (along the coastline) between the Maui A platform and the pumping station, then the situation can be represented as:

9 Maui A Maui B 40 0 Pumping Station 80 - Using the Pythagorean Theorem the total length of the pipeline required to connect both platforms to the pumping station can be epressed as a function of : Differentiating with respect to : L( ) = ( 80! ). L!( ) = ( ) ( ) " 80 " " Setting the derivative equal to zero and solving for gives: = 60/ km. Therefore, in order minimize the total length of the two pipelines, the pumping station should be built 60/ km from the Maui A platform (distance measured along the Taranaki coast).