The dimensions of a rectangle with perimeter 36 cm have to be determined that maximize the area without using calculus.

Let the length of the rectangle be L and the width be W. The perimeter of a rectangle is 2(L + W) = 36

=> W = 18 - L

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The dimensions of a rectangle with perimeter 36 cm have to be determined that maximize the area without using calculus.

Let the length of the rectangle be L and the width be W. The perimeter of a rectangle is 2(L + W) = 36

=> W = 18 - L

The area of the rectangle is A = L*W = L*(18 - L)

=> 18L - L^2

The graph of A = 18L - L^2 has been plotted below with area on the y-axis and length L on the x-axis

As can be seen the graph peaks at A = 81. At A = 81, L = 9

The length of the required rectangle has a length equal to 9 and the width is 18 - 9 = 9.

**The required rectangle is a square with sides equal to 9 cm and the maximum area is 81 cm^2.**