A rectangle has a perimeter of 32 in. find the length and width of the rectangle under which the area is the largest. follow the steps: (a) let the width to be x and the length to be y, then the quantity to be maximized is (expressed as a function of both x

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15 October, 14:09

A rectangle has a perimeter of 32 in. find the length and width of the rectangle under which the area is the largest. follow the steps: (a) let the width to be x and the length to be y, then the quantity to be maximized is (expressed as a function of both x

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June Ritter · Answer 1

Let width and length be x and y respectively.

Perimeter (32in) = 2x+2y=> 16=x+y = > y=16-x

Area, A = xy = x (16-x) = 16x-x^2

The function to maximize is area: A=16 x-x^2

For maximum area, the first derivative of A = 0 = > A'=16-2x = 0

Solving for x: 16-2x=0 = >2x=16 = > x=8 in

And therefore, y=16-8 = 8 in