MATH 12002 - CALCULUS I 3.5: Optimization (Part 3) Professor Donald - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §3.5: Optimization (Part 3)

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Example 1

Example 1

A box with an open top is to be constructed from a square piece of cardboard, 3 feet wide, by cutting out a square from each of the corners and bending up the sides. Find the largest volume that such a box can have.

3 3

x x

3 3

x x 3 − 2x ✲

✛

3 − 2x

✻ ❄

With x equal to the length of a side of the cut-out square, the volume is V (x) = x(3 − 2x)2 and 0 x 1.5.

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Example 1

We want to find the absolute maximum value of V (x) = x(3 − 2x)2

• n the closed interval [0, 1.5].

We need the derivative in order to find any critical numbers of V (x): V ′(x) = 1 · (3 − 2x)2 + x · 2(3 − 2x)1 · (−2) = (3 − 2x)[(3 − 2x) + x · 2 · (−2)] = (3 − 2x)[(3 − 2x) − 4x] = (3 − 2x)(3 − 6x). The critical numbers of V (x) are then x = 3/2 = 1.5 and x = 1/2 = 0.5. Evaluating V (x) at the critical numbers and endpoints, we have V (0) = 0 · (3 − 2(0))2 = 0 V (1.5) = 1.5 · (3 − 2(1.5))2 = 1.5 · (3 − 3)2 = 0 V (0.5) = 0.5 · (3 − 2(0.5))2 = 0.5 · (3 − 1)2 = 2, and the largest possible volume is 2 cubic feet.

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Example 2

Example 2

A llama farmer wishes to enclose a rectangular field of 20000 square feet with a fence. For the side running along the road, he will use fencing that costs $3 per foot and for the other 3 sides he will use fencing that costs $1 per foot. What dimensions will minimize the cost of the fence?

ROAD

x y x y Let x = length of the side perpendicular to the road; y = length of the side parallel to the road. Given: Area = xy = 20000 square feet. Minimize: Cost C of fence.

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Example 2

The fence has two sides, each of length x, that cost $1 per foot.

• ne side of length y that costs $1 per foot;
• ne side of length y that costs $3 per foot;

Hence the cost of the fence is C = x + x + y + 3y = 2x + 4y. Using the constraint xy = 20000, so y = 20000/x, we have C(x) = 2x + 4 · 20000 x = 2x + 80000x−1.

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Example 2

We want to find the absolute minimum value of C(x) = 2x + 80000x−1

• n (0, ∞).

We need the derivative in order to find any critical numbers of C(x): C ′(x) = 2 − 80000x−2 = 2 − 80000 x2 = 2x2 − 80000 x2 = 2(x2 − 40000) x2 = 2(x − 200)(x + 200) x2 .

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Example 2

Since C ′(x) = 2 − 80000x−2 = 2(x − 200)(x + 200) x2 , the only critical number on the interval (0, ∞) is x = 200. Verify that C(x) has an absolute minimum at x = 200: C ′′(x) = 160000x−3, so C ′′(200) = 160000/2003 is positive. By the Second Derivative Test, C has a local minimum at x = 200. The domain of C is an interval and C has only one critical number, hence C has an absolute minimum at x = 200. Therefore, the cost of the fence is minimized when x = 200 feet and y = 20000/200 = 100 feet. The dimensions are then 200 ft × 100 ft, where the short side is along the road.

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