Question

Solve the following two problems using Calculus. Show all of your work on a separate sheet of paper, making clear how you arrived at all your solutions. Please be aware that you are using similar thinking for both of these problems, but you are solving for maximum area in Part A and minimum cost in Part B. Enjoy

A. If the materials for the fence cost $12 per foot, find the dimensions of the
corral for the largest possible area that can be enclosed with $2400 worth of fence.

B. If he only requires an area of 288 square feet for his vegetable garden, find the minimum cost of putting up this fence, if the cost per foot is $10. Indicate the dimensions of the garden that will minimize the cost.​

Answer 1

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Answer:

A. 50 ft square

B. 12√2 ft square

Explanation:

These problems are basically the same, so have the same solution. The rectangle with maximum area for a given perimeter will have the same shape as the one with minimum perimeter for a given area.

We can find it generically. Let p represent the perimeter of a rectangular area with one side that measures x. The area will be ...

A = x(p/2 -x) = -x^2 +(p/2)x

The area will be maximized when dA/dx = 0:

dA/dx = -2x +p/2 = 0

p/2 = 2x . . . . . add 2x

x = p/4 . . . . . . divide by 2

The other dimension is ...

p/2 -x = p/2 -p/4 = p/4

The dimensions of the maximum area for perimeter p are ...

p/4 × p/4 . . . . . . . a square

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A.

$2400 worth of fence at $12 per foot is 2400/12 = 200 ft of fence. The largest possible area that can be enclosed will have dimensions of 200 ft/4 = 50 ft square:

50 ft × 50 ft

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B.

The perimeter of the garden will be at its shortest when the shape of the garden is square.

√(288 ft²) = 12√2 ft

The dimensions of the garden that will minimize the cost are ...

12√2 ft × 12√2 ft