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What is the maximum area for a rectangle whose perimeter is 48 meters?

User Mithat
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1 Answer

5 votes

Answer:

12m²

Explanation:

For a rectangle, with length L and width W,

the perimeter is given as

Perimeter,

P = (2 x Length) + (2 x Width)

P = 2L + 2W

It is given that the perimeter is 48, hence

48 = 2L + 2W (divide both sides by 2)

24 = L + W

or

L = 24 - W -----> eq 1

Also realize that the Area of a Rectangle is given by

A = L x W -----> eq 2

Substituting eq 1 into eq 2,

A = (24 - W) x W

A = -W² + 24W

Recall that for a quadratic equation y = ax² + bx + c, the maxima or minima is given by y(max) = -b/2a

In this case, b = 24 and a = -1

-b/2a = -24/[ 2(-1) ] = 12

Hence for A to be maximum A(max) = 12m² (Answer)

User Jhericks
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