9514 1404 393
Answer:
A. 50 ft square
B. 12√2 ft square
Step-by-step 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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<h3>A. </h3>
$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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<h3>B.</h3>
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
