Optimization Practice Solutions, introduction to calculus, PDF

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This document contains solutions to "Optimization Practice, introduction to calculus,basic problems" for calculus intro course Math111 at Emory University....

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Optimization 1. Your goal on Thanksgiving is to eat as much as possible without passing out. If you eat politely, making each forkful 1 gram of food, you can eat 200 forkfuls of food before you’re completely stuffed. If you eat more per forkful, the number of forkfuls you can eat decreases: for each gram you increase your forkful, you can eat 8 fewer forkfuls. What is the largest amount of food you can eat? Solution. We’ll follow the steps outlined on the worksheet. (a) Goal: Maximize the amount of food you can eat. (b) Name variables: Let’s make a variable x which represents the size of each forkful (in grams) and a variable y which represents the number of forkfuls you can eat. (c) What are you optimizing?: We want to maximize the amount of food you can eat, which is equal to the product (number of forkfuls you can eat) (size of each forkful), or xy. Therefore, we’re trying to optimize the product xy . (d) Reduce to one variable: We know that, when x = 1, y = 200. We’re told that, each time we increase x by 1, y decreases by 8. This tells us that y is a linear function of x with a slope of −8. So, y = −8x + b for some b. We can solve for b by plugging in x = 1, y = 200, which gives b = 208. So, y = −8x + 208. We were trying to maximize xy, which we now know is x(−8x + 208). (e) Find the domain: The smallest x could be is 1 (we have to eat something ). The biggest x could be is 26 (once x reaches 26, you can’t eat any forkfuls of food). So, our problem is now: “Maximize f (x) = x(−8x + 208) on the interval [1, 26].” (f) Find the extrema: We first find the critical points of f . Since f (x) = −8x2 + 208x, f ′ (x) = −16x + 208, which is 0 when x = 13. So, we should test 1, 13, and 26: f (1) = 200, f (13) = 1352, and f (26) = 0. By the Closed Interval Method, the maximum occurs when x = 13. (g) Answer the original question: The question asks for the largest amount of food we can eat, which is f (13) = 1352 grams . 2. You need to make a poster for your CS170 class. Davide insists that your poster has to be 200 square inches and will have 1 inch margins on the sides, a 3 inch margin on the bottom and a 1 inch margin on the top. What dimensions will give the largest printed area? Solution. (a) Goal: Maximize the area of the printed area. (b) Name variables: Below are some sketches of possible poster sizes with the margins drawn. On the first picture, we have labeled the dimensions of the poster with x and y.

1

x

y

(c) What are you optimizing?: The area of the printed region is not xy. We must adjust each length by the desired margins. The 1 inch margin at the top and 3 inch margin at the bottom mean that y − 1 − 3 = y − 4 is the vertical length we have to work with. Similarly, the 1 inch margin on the sides means we actually have x − 1 − 1 = x − 2 inches horizontally to work with. Therefore, area of the printed region is A = (x − 2)(y − 4), which is what we’re trying to optimize. (d) Reduce to one variable: We want to find a single variable function. To accomplish this we realize that since the poster has to have an area of 200 square inches there is a relationship between the . We can use this to find the area of the printed two variables: xy = 200. Therefore, x = 200  y  200 region as a function of one variable: A = y − 2 (y − 4), whence A = 208 − 2y − 800 . y We were trying to maximize (x − 2)(y − 4), which we now know is 200 − 2y −

800 y

+ 8.

(e) Find the domain: Since length can’t be negative, and the printed area inside has dimensions x − 2 and y − 4, we need to have y − 4 ≥ 0, whence y ≥ 4. Similarly, we need x − 2 ≥ 0, whence x ≥ 2. ≥ 2. Solving Since xy = 200, we also have x = 200 , so this last inequality can be rewritten as: 200 y y for y we obtain y ≤ 100. Thus, the domain of this function is [4,100]. So, our problem is now: “Maximize A(y) = 208 − 2y −

800 y

on the interval [4, 100].”

(f) Find the extrema: We first find the critical points of A. 800 y2

A′ = −2 + 0 = −2 +

800 y2

2y2 = 800 y2 = 400 y = 20 To determine the nature of this critical point, we can make use of the second derivative test A′′ = −

1600 y3

A′′ (20) < 0 2

This means that at y = 20 we have a local maximum. Since there is only one critical point, this local max is also a global maximum. Note that we could have checked the endpoints here as well, and done the Closed Interval Method. We would get that A(4) = A(100) = 0. Thus, y = 20 is indeed the absolute maximum. (g) Answer the original question: If y = 20, then x = printed area are: x=10,y=20 .

200 20

= 10. Thus, the dimensions for the largest

3. Walter is planning to buy a custom-made jewelry box for his mother’s next birthday. The box will have a square base. The sides and bottom will be made out of mahogany, which costs 30 cents per square inch. The top will be made out of maple, which costs 50 cents per square inch. Walter has $60 to spend on the present and wants to get a box with the largest volume possible. What dimensions should his box be? Solution. (a) Goal: Maximize the volume of the box. (b) Name variables: This box has a square base. Let s be the length of the sides of the square and h be the height of the box. Let V be the volume of the box. (c) What are you optimizing?: The volume of the box is given by the area of the base times the height. That is, V = s2 h. (d) Reduce to one variable: We want to express V as a function of just one variable. To do this, we need to relate s and h using something we know. We are told that the cost of the box is $60. Let’s work in cents: 6000 = cost of box (in cents) = cost of top + cost of bottom + 4(cost of each side) The top costs 50 cents per square inch, while all other sides cost 30 cents per square inch: = (50)(area of top) + (30)(area of bottom) + 4(30)(area of each side) = 50s2 + 30s2 + 4(30)sh = 80s2 + 120sh Now, we can solve this equation for h: 6000 = 80s2 + 120sh Divide both sides by 40 to make the numbers simpler: 150 = 2s2 + 3sh 150 − 2s2 = 3sh

150 − 2s2 =h 3s 50 2s − =h s 3 Plugging this into our volume formula gives   50 2s 2s3 2 V =s − = 50s − 3 s 3 3

(e) Find the domain: Note that s must be positive because it represents a length. Also, the volume 3 3 must be positive as well. If we set up V > 0 we have 50s − 2s3 > 0, whence 50s > 2s3 . Dividing √ by s (which is positive) and clearing denominators we’re left with 75 > s2 , whence s < 5 3. √ Therefore, s must be in the interval (0,5 3). So, our problem is now: “Maximize V (s) = 50s −

2s3 3

√ on the interval (0,5 3).”

(f) Find the extrema: We want the critical points, so we should first find the derivative of V (s), which is V ′ (s) = 50 − 2s2 . So, the critical points are where 50 − 2s2 = 0 (V ′ (s) is never undefined) 50 = 2s2 25 = s2 5=s (s must be positive because it represents a length.) But wait, we’re not done yet! We’ve found a critical point, but we still need to check that it’s the absolute maximum. Let’s use the Second Derivative Test: V ′′ (s) = −4s, so V ′′ (5) = −20 < 0. Therefore, 5 is a local maximum. Since it’s the only critical point in our domain, it must be an absolute maximum as well. So, s = 5 is an absolute maximum. 50 2s − , the height when s = 5 is s 3 20 20 50 2 · 5 = . So, the dimensions for the box should be 5 inches × 5 inches × − inches . 3 3 5 3

(g) Answer the original question: Since we found that h =

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