What is the max. area that can be created if a rectangular cage is to be made with a wall as one side and 80 meters of fencing for the other three sides?

Answer 1

Let the length (parallel to the wall) be L and the width (perpendicular to the wall) be W, then:

The Fencing used along three sides is:

L + 2W = 80

→ L = 80 - 2W

And the area of the enclosed rectangle is:

area = LW

→ area = (80 - 2W)W

→ area = 80W - 2W²

Completing the square gives:

→ area = -2(W² - 40)

→ area = -2((W - 20)² - 20²)

→ area = 800 - 2(W - 20)²

As a square is always positive, this has a maximum value of 800 when W - 20 = 0

→ max area is 800 m² (and occurs when W = 20m and L = 80 - 2W = 40m).

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Alternatively, using calculus:

for a maximum area: d/dW area = 0

→ d/dW 80W - 2W² = 80 - 4W = 0

→ 4W = 80

→ W = 20

→ L = 80 - 2W = 80 - 2×20 = 40

→ max_area = 40×20 = 800 m²

(Note that the non-calculus method of completing the square gives the maximum area without the need to actually work out the dimensions of the enclosure, though for the calculus method, once the width is known the area can be calculated directly: area = 80×20 - 2×20² = 800 m² but that involves large numbers, unless it is calculated as 20(80 - 2×20) which is effectively calculating the length.)