MATH 12002 - CALCULUS I 3.5: Optimization (Part 1) Professor Donald - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §3.5: Optimization (Part 1)

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Example 1

Example 1

A rectangular cabbage patch 200 square feet in area is to be laid out so that an adjacent straight wall serves as one of the sides. What are the dimensions of the patch requiring the least amount of fence?

WALL ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣ ♣

x y x Let x = length of side perpendicular to the wall; y = length of side parallel to the wall. Given: Area = xy = 200 square feet. Minimize: Length of fence L = 2x + y.

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Example 1

We can write the length of the fence as a function of one variable, x, by using the constraint that xy = 200, hence y = 200/x. Substituting in the formula L = 2x + y, we have L(x) = 2x + 200 x = 2x + 200x−1. We want to find the absolute minimum value of L(x) on (0, ∞). We need the derivative in order to find any critical numbers of L(x): L′(x) = 2 − 200x−2 = 2 − 200 x2 = 2x2 − 200 x2 = 2(x2 − 100) x2 = 2(x − 10)(x + 10) x2 .

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Example 1

Since L′(x) = 2 − 200x−2 = 2(x − 10)(x + 10) x2 , the only critical number on the interval (0, ∞) is x = 10. Verify that L(x) has an absolute minimum at x = 10: L′′(x) = 400x−3 = 400/x3, so L′′(10) = 400/103 is positive. By the Second Derivative Test, L has a local minimum at x = 10. The domain of L is an interval and L has only one critical number, hence L has an absolute minimum at x = 10. Therefore, the length of the fence is minimized when x = 10 feet and y = 200/10 = 20 feet. The dimensions are then 10 ft × 20 ft.

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Example 2

Example 2

A closed box is to be made out of heavy cardboard. If the base of the box is a rectangle twice as long as it is wide, and the box is to hold 9 cubic feet, what dimensions require the least amount of cardboard?

✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ w ℓ = 2w h

Let w = width, ℓ = length, and h = height of the box. Given: Volume = ℓ × w × h = 2w2h = 9 cubic feet. Minimize: Surface area A of the box.

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Example 2

We first need a formula for A in terms of our variables.

✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ w 2w h

The box is to have a top, so there are two sides, each with area 2wh; a front and back, each with area wh; and a bottom and top, each with area 2w2. Hence A = 2(2wh) + 2(wh) + 2(2w2) = 4w2 + 6wh. Using the constraint 2w2h = 9, so h = 9/(2w2), we have A = 4w2 + 6w · 9 2w2 = 4w2 + 27 w .

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Example 2

As a function of w, the surface area is A(w) = 4w2 + 27 w = 4w2 + 27w−1. We want to find the absolute minimum value of A(w) on (0, ∞). We need the derivative in order to find any critical numbers of A(w): A′(w) = 8w − 27w−2 = 8w − 27 w2 = 8w3 − 27 w2 . The only critical number of A(w) in (0, ∞) is where 8w3 − 27 = 0, hence w =

3

• 27

8 = 3 2 = 1.5.

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Example 2

Verify that A(w) has an absolute minimum at w = 1.5: A′(w) = 8w − 27w−2, so A′′(w) = 8 + 54w−3 = 8 + 54/w3 and A′′(1.5) = 8 + 54/(1.53) is positive. By the Second Derivative Test, A has a local minimum at w = 1.5. The domain of A is an interval and A has only one critical number, hence A has an absolute minimum at w = 1.5. Therefore, the amount of cardboard required is minimized when w = 1.5 feet, ℓ = 2(1.5) = 3 feet, and h = 9/(2(1.52)) = 2 feet. The dimensions are then 1.5 ft × 3 ft × 2 ft.

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