Ma-min Word Problems In this section, we ll use our results on maima and minima for functions to do word problems which involve finding the largest or smallest value of lengths, areas, volumes, costs, and so on. The hardest part of doing these problems is setting up the appropriate equations; the calculus part is relatively simple. Some of these problems involve finding an absolute ma or min on a closed interval. We know that to do this, we find the critical points on the interval, then test the endpoints and the critical points in the original function. In some situations, the quantity we re trying to maimize or minimize varies over a non-closed interval. In those cases, the following theorem is often useful. Theorem. Suppose a function f has its second derivative f defined on an interval a < < b. Suppose that f (c) = 0, and that c is the only critical point on the interval a < < b. (a) If c is a local ma, then it is an absolute ma. (b) If c is a local min, then it is an absolute min. The key in applying the theorem is that there is only one critical point. Eample. A rectangular pen consists of two identical sections separated by a fence. 10 feet of fence is used for the outside of the pen and the dividing fence. What dimensions for each section produce the pen of largest area? Let each section be feet in width and y feet in height. y y y The total area is A = y. Since 10 feet of fence are available, 10 = 4+3y y = 1 3 (10 4) Plug this into A = y: A = 1 3 (10 4) = 3 (10 4 ). The endpoints are given by the etreme cases = 0 and y = 0; note that y = 0 gives 10 = 4, or = 30. Differentiate: A = 3 (10 8). 1
Set A = 0 and solve for : 3 (10 8) = 0 = 15 0 15 30 A 0 600 0 The maimum occurs when = 15. In this case, y = 1 (10 4 15) = 0. 3 Eample. A rectangular sheet of cardboard is 15 wide and 4 inches high. Squares with sides of length are cut out of each corner. The four resulting tabs are folded up to make a rectangular bo (with no top). Find the value of which gives the bo of largest volume. 4 15 15-4- The base of the bo is 15 by 4, and the height is. Hence, the volume is V = (15 )(4 ) = 4 3 78 +360. The endpoints are = 0 (cut no squares out) and = 15 (cut halfway from the bottom to the top). Obviously, these endpoints give volume 0, but it means that V is defined on a closed interval. So I just need to test the critical points on the interval and find the one that gives a ma. Differentiate: V = 1 156+360 = 1( 13+30) = 1( 3)( 10). The roots are = 10 and = 3. Now = 10 is greater than 15, so it s outside the interval. So I only need to consider the other root = 3. Thus, = 3 maimimizes the volume. 0 3 15 V 0 486 0 Eample. A rectangular poster is printed on a piece of cardboard with area 100 square inches. The printed region is a rectangular region centered on the cardboard. There are unprinted (blank) margins of width 3
inches on the left and right of the printed region, and width 4 inches on the top and bottom of the printed region. What dimensions for the printed region maimize its area? +6 4 y printed 3 region 3 y+8 margin 4 Suppose the printed area is by y. Its area is A = y. Since there are margins of width 3 on the left and right, and margins of width 4 on top and bottom, the area of the cardboard is (+6)(y +8) = 100. Solving for y yields Plug this into the epression for A to obtain y = 100 +6 8. ( ) 100 A = +6 8 = 100 +6 8. The etreme cases are = 0 and y = 0. If y = 0, then (+6)(8) = 100 = 144 So the endpoints are = 0 and = 144. (These make the area of the printed region 0, so they obviously don t give the maimum!) Compute the derivative: A = (+6)(1) (100)(1) (+6) 8 = 700 (+6) 8. Find the critical points: 700 (+6) 8 = 0 700 (+6) = 8 700 = 8(+6) 900 = (+6) ±30 = +6 6±30 = 3
Since is a length, it can t be negative, so take the plus sign. This gives Plug this into y = 100 8 to get y = 3. +6 = 30 6 = 4. 0 4 144 A 0 768 0 This shows that = 4 and y = 3 maimize the area of the printed region. Eample. Find the point on the line 8y +3 = 0 which is closest to the point (1,6). The distance from (,y) to (1,6) is d = ( 1) +(y 6). (1,6) (,y) 8y - + 3 = 0 I want the point (,y) for which the distance is smallest. Since smaller numbers give smaller squares and vice versa, I can find where the square of the distance is smallest: s = ( 1) +(y 6). This allows me to remove the square root, and will make differentiation easier. The line is 8y +3 = 0. Solving for gives = 8y +3. Plug this into the equation for s: S = [(8y +3) 1] +(y 6) = (8y +) +(y 6). There are no restrictions on, so I ll use the Second Derivative Test. Differentiate (noting that the variable is y): s = (8y +)(8)+(y 6) = 18y +3+y 1 = 130y +0, Set s = 0 to find the critical points: s = 130. 130y +0 = 0 130y = 0 y = 13 This gives ( = 8 ) +3 = 3 13 13. 4
min. ( Now s ) = 130 > 0, so y = is a local min. Since it s the only critical point, it s an absolute 13 13 Eample. The volume of a circular cylinder (with a top and a bottom) is 000π. What values for the radius r and the height h give the smallest total surface area (the area of the side plus the area of the top plus the area of the bottom)? r h The area of the side is πrh, the area of the top is πr, and the area of the bottom is πr. The total surface area is A = πrh+πr. The volume is Solving for h gives h = 000. Plug this into A: r 000π = πr h, so 000 = r h. A = πr 000 r +πr = 4000π +πr. r The only restriction on r is r > 0. Hence, r is not restricted to a closedinterval a r b. Therefore, I ll use the Second Derivative Test. Compute A and A : A = 4000π r +4πr, Set A = 0 and solve for r: This gives h = 000 100 = 0. Now A = 8000π r 3 +4π. 4000π r +4πr = 0 A (k) = 8000π 1000 4πr 3 = 4000π r 3 = 1000 r = 10 +4π = 1π > 0. Hence, r = 10 is a local min. Since it s the only critical point, it s an absolute min. 5
Eample. A cylindrical can with a top and a bottom is made with 1536π square inches of sheet metal, with no waste. What dimensions for the radius r and the height h give the can of largest volume? r h V = πr h. The area of the top is πr, the area of the bottom is πr, and the area of the side is πrh. So the total area is 1536π = πr +πrh. Solve the equation for h: Substitute in V: V = πr 768 r r 1536π = πr +πrh 768 = r +rh 768 r = rh 768 r = h r = πr(768 r ) = π(768r r 3 ). r = 0 is ruled out, because it would cause division by 0 in the equation for h. There is no other restriction on r, ecept that it should be positive. Since V is not restricted to a closed interval [a,b], I ll use the Second Derivative Test. Compute the derivatives: V = π(768 3r ), Find the critical points: V = π( 6r) = 6πr. π(768 3r ) = 0 768 3r = 0 768 = 3r 56 = r ±16 = r Since r can t be negative (as it s the radius of a cylinder), I get r = 16. Then The second derivative is h = 768 56 16 = 3. V (k) = 6π 16 = 96π < 0. Thus, r = 16 is a local ma. Since it s the only critical point, it s an absolute ma. 6
Eample. Find the positive number for which the sum of 5 times the number and 30 times its reciprocal is smallest. Let be the number. The sum of 5 times the number and 30 times its reciprocal is S = 5+30 1 30 = 5+. The only restriction on is > 0. Since is not restricted to a closed interval [a,b], I ll use the Second Derivative Test. Differentiate: S = 5 30, Find the critical points by setting S = 0: S = 640 3. 5 30 = 0 5 = 30 5 = 30 = 8 = ±8 Since > 0, I get = 8. Plug this into the Second Derivative: S = 640 8 = 10 > 0. Hence, = 8 is a local min. Since it s the only critical point, it s an absolute min. Eample. A rectangular bo with a square bottom and no top is made with 100 square inches of cardboard. What values of the length of a side of the bottom and the height y give the bo with the largest volume? y The volume is V = y. The area of the 4 sides is 4y, and the area of the bottom is. So 100 = 4y +. 7
Solving for y gives Plug this into V and simplify: V = 100 4 y = 100. 4 = 1 4 (100 ) = 1 4 (100 3 ). Note that 0, since = 0 plugged into 100 = 4y + gives 100 = 0 a contradiction. So the only restriction on is that > 0. Since is not restricted to a closed interval [a,b], I ll use the Second Derivative Test. Compute the derivatives: V = 1 4 (100 3 ). V = 1 4 ( 6) = 3. Find the critical points: 1 4 (100 3 ) = 0 100 3 = 0 3 = 100 = 400 = ±0 Since is a length, it must be positive, so = 0. This gives Plug = k into the Second Derivative: y = 100 400 80 = 10. V (k) = 3 0 = 30 < 0. = 0 is a local ma, but it s the only critical point, so it s an absolute ma. Eample. A rectangular bo with a square bottom and no top has a volume of 048 cubic inches. What values of the length of a side of the bottom and the height y give the bo with the smallest total surface area (the area of the bottom plus the area of the sides)? y The area of the 4 sides is 4y, and the area of the bottom is. So the total area is A = 4y +. 8
The volume is Solving for y gives Plug this into A and simplify: 048 = y. y = 048. A = 4 048 + = 819 +. Note that 0, since = 0 plugged into 048 = y gives 048 = 0, a contradiction. So the only restriction on is that > 0. Since is not restricted to a closed interval [a,b], I ll use the Second Derivative Test. Compute the derivatives: A = 819 +, Find the critical points: = 16 gives Plug = 16 into the Second Derivative: A = 16384 3 +. 819 + = 0 = 819 3 = 819 3 = 4096 = 16 y = 048 56 = 8. A (16) = 16384 4096 + = 6 > 0. = 16 is a local min, but it s the only critical point, so it s an absolute min. Eample. A rectangular bo is made with cardboard. It has no top, and consists of two identical partitions separated by a common wall. Each partition has a square bottom which is inches by inches, and the height of the bo is y inches. If the volume of the whole bo is 10584 cubic inches, what values for and y minimize the total amount of cardboard used (for the bottom, the four sides, and the divider that separates the partitions)? y 9
The area of the bottom is. The front and back are each by y, so they have a total area of y = 4y. The right side, left side, and the middle partition are each by y, so they have a total area of 3y. Therefore, the total area (the total amount of cardboard used) is A = +4y +3y = +7y. The total volume is Solving this equation for y gives Plug this into A and simplify: 10584 = y. y = 59. A = +7 59 = + 37044. The etreme case = 0 is ruled out, because it causes division by 0 in the equation for A. Thus, is not restricted to a closed interval [a,b], and I ll use the Second Derivative Test. Differentiate: A = 4 37044, A = 4+ 74088 3. Find the critical points by setting A = 0 and solving for : 4 37044 = 0 4 = 37044 4 3 = 37044 3 = 961 = 1 This gives Do the Second Derivative Test: y = 59 441 = 1. V (1) = 4+ 74088 961 = 1 > 0. Therefore, = 1 is a local min. Since it s the only critical point, it s an absolute min. Eample. A rectangular bo with no top has two identical partitions separated by a common wall. Each partition has a square bottom. If 400 square inches of cardboard are used with no waste to construct the bo, what dimensions yield the bo with the largest (total) volume? 10
y Suppose the base of each partition is by and the height is y. The volume is V = y. The area of the bottom is, the area of the front is y, the area of the back is y, the area of the left side is y, the area of the right side is y, and the area of the divider that separates the two partitions is y. The total area is +y +y +y +y +y = 400, so +7y = 400. Solving for y yields y = 400. Plug this into the volume equation and simplify: 7 V = 400 7 = 7 (400 ) = 7 (400 3 ). can t be 0, since = 0 causes division by 0 in the formula for y. Since is a length, it must be positive. So > 0, and there is no other restriction on. Since is not restricted to a closed interval [a,b], I ll use the Second Derivative Test. Differentiate: V = 7 (400 6 ), Find the critical points by setting V = 0: V = 7 ( 1) = 4 7. 7 (400 6 ) = 0 Since must be positive, I get = 0. This gives 400 6 = 0 400 = 6 400 = ±0 = y = 400 800 140 = 80 7. Plug = 0 into V : V = 4 0 = 480 7 7 < 0. Hence, = 0 is a local ma. Since it s the only critical point, it s an absolute ma. c 018 by Bruce Ikenaga 11