# Homework from Section Find two positive numbers whose product is 100 and whose sum is a minimum.

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1 Homework from Section Find two positive numbers whose product is 100 and whose sum is a minimum. We want x and y so that xy = 100 and S = x + y is minimized. Since xy = 100, x = 0. Thus we have y = 100/x, so we want to minimize S(x) = x + 100/x. We find that S (x) = 1 100/x 2. Setting S (x) = 0 and solving for x yields x = ±10. Since we are looking for positive numbers, we take x = 10. Notice that for 0 < x < 10, x 2 < 100, so 100/x 2 > 1, which implies that 1 100/x 2 < 0. Thus S (x) < 0 for such x. Similarly, for x > 10, S (x) > 0. It follows from the first derivative test that x = 0 is a minimum of S(x). Thus x = 10 and y = 100/x = 10 is a pair of positive numbers whose product is 100 and whose sum is a minimum Consider the following problem: A farmer with 750 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens? (a) Draw several diagrams illustrating the situation, some with shallow, wide pens and some with deep, narrow pens. Find the total areas of these configurations. Does it appear that there is a maximum area? If so, estimate it. (b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with your symbols. (c) Write an expression for the total area. (d) Use the given information to write an equation that relates the variables. (e) Use part (d) to write the total area as a function of one variable. (f) Finish solving the problem and compare the answer with your estimate in part (a). (a) Here are three diagrams. The area of the one on the left is ft 2, the area of the middle one is ft 2, and the area of the one on the left is 9000 ft 2. It seems that there is a maximum and I would put it at around ft 2. 1

2 (b) Let x be one side of the overall pen and y the other. The general situation is (c) The total area is A = xy. (d) The variables are related by 5x + 2y = 750. (e) Since 5x + 2y = 750, we have y = x. Thus we can write A(x) = xy = 375x 5 2 x2. (f) We find that A (x) = 375 5x. Setting A (x) = 0, we solve to find x = 75. Note that for x < 75, A (x) > 0 and for x > 75, A (x) < 0. Thus x = 75 is a maximum of A by the first derivative test. Furthermore, A(75) = ft 2, which is larger than my estimate in part (a) (a) Show that of all rectangles with a given area, the one with smallest perimeter is a square. (b) Show that of all rectangles with a given perimeter, the one with greatest area is a square. (a) Let x and y be the sides of the rectangle. We have A = xy, where A is constant. We want to minimize P = 2x + 2y. Since xy = A, we have y = A/x. Thus we have P(x) = 2x + 2A/x. Setting P (x) = 0, we have 0 = 2 2A/x 2. Thus x = ± A. Since side lengths of rectangles are positive, we have x = A. Note that for 0 < x < A, P (x) < 0 and for x > A, P (x) > 0. The argument for this is similar to that in problem Thus x = A is a minimum of P. Furthermore, we have y = A/x = A/ A = A = x. Since x = y in this case, the rectangle with smallest perimeter and area A is a square. (b) Let x and y be the sides of the rectangle. We have P = 2x + 2y where P is constant. Hence we have y = 1 2 P x. We want to maximize A = xy. Plugging in for y we have A(x) = P 2 x x2. Thus A (x) = P/2 2x. Setting P (x) = 0 and solving for x we get x = P/4. Note that for 0 < x < P/4, A (x) > 0 and for x > P/4, A (x) < 0. Thus, by the first derivative 2

3 test, x = P/4 is a maximum. In this case we have y = P/2 x = P/2 P/4 = P/4 = x. Since x = y we have that of all rectangles with a given perimeter, the one with greatest area is a square Find the points on the ellipse 4x 2 + y 2 = 4 that are farthest away from the point (1, 0). We want to maximize d = (x 1) 2 + y 2 given that 4x 2 + y 2 = 4. Notice that to have 4x 2 + y 2 = 4, we must have x 1. While we could solve for x or y and plug it in, it is better not to in this case because we do not want to deal with taking plus or minus square roots. Rather, notice that d = (x 1) 2 + y 2 = x 2 2x y 2 = 1 2x 3x 2 + 4x 2 + y 2 = 1 2x 3x = 5 2x 3x 2. Thus we want to maximize d(x) = 5 2x 3x 2. Setting the derivative equal to 0 yields 0 = d (x) = 2 6x 2 5 2x 3x 2. Solving yields x = 1/3. For 1 < x < 1/3, 2 6x > 0, so d (x) > 0. Similarly, for 1/3 < x < 1, we have d (x) < 0. Thus x = 1/3 is a maximum of d by the first derivative test. Solving 4x 2 + y 2 = 4 for y, we find that y = ±2 1 x 2, so at x = 1/3, y = ±4 2/3. Thus the two points on the ellipse 4x 2 + y 2 = 4 that are farthest away from the point (1, 0) are ( 1/3, 4 2/3) and ( 1/3, 4 2/3) Find the area of the largest rectangle that can be inscribed in the ellipse x 2 /a 2 + y 2 /b 2 = 1. In this case the area of the rectangle is A = 4xy, where (x, y) is the coordinate of the corner in the first quadrant. We may assume a, b > 0. Solving x 2 /a 2 + y 2 /b 2 = 1 for y, we get y = ±b 1 x 2 /a 2. Since y lies in the first quadrant we have y = b 1 x 2 /a 2. Thus we have A(x) = 4bx 1 x 2 /a 2. Setting A (x) = 0 we find Multiplying by a 2 1 x 2 /a 2 gives 0 = 4b 1 x 2 /a 2 4bx 2 a 2 1 x 2 /a 2. 0 = 4ba 2 (1 x 2 /a 2 ) 4bx 2 = 4ba 2 8bx 2, so x = a/ 2, where again we have chosen the positive value of the square root. Analysis like those in previous problems shows that A is maximized by this value of A. Thus the maximum area is A(a/ 2) = 2ba. 3

4 A cone shaped drinking cup is made by cutting out a sector of a paper circle of radius R and joining the two edges. Find the maximum capacity of such a cup. Let r be the radius of the circle at the top of the cone shaped cup. The Pythagorean theorem tells us that the height of the cone is h = R 2 r 2. Thus the volume is Setting the derivative equal to 0 we get V(r) = π 3 r2 h = π 3 r2 R 2 r 2. 0 = V (r) = 2πr R 2 r 2 3 πr 3 3 R 2 r 2. Notice that r = 0 is a solution, but in this case V(r) = 0, which cannot be the maximum (also, 0 is not in the domain of V ). Solving with r = 0 yields the positive solution r = 6R/3. Analysis similar to previous problems shows this is a maximum. Thus the maximum capacity is V( 6R/3) = 2πR Find the equation of the line through (3, 5) that cuts off the least area from the first quadrant. Using the point slope form, the general equation for the line is y = m(x 3) + 5. If m 0, this line cuts off no area from the first quadrant, but this is a triviality, so assume m < 0. Then we x intercept is at x = 3 5/m and the y intercept is at y = 5 3m. Notice that the line cuts of a triangle with base the distance from the origin to the x intercept and height the distance to the y intercept. Thus the area cut off is A(m) = xy/2 = (5 3m)(3 5/m)/2 = 1 (30 25/m 9m). 2 Setting A = 0, we get 0 = 25/m 2 9. Solving for m we get m = 5/3, where we take the negative value of the square root since we assumed m < 0. The first derivative test shows this m minimizes A, so the equation of the line through (3, 5) that cuts off the least area from the first quadrant is y = 3 5 x

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