Find the dimensions of a rectangle with perimeter 68 m whose area is as large as possible. (If both values are the same number, enter it into both blanks.)

The dimensions are 17m and 17m.

The perimeter of a rectangle is given as:

= 2(length + width)

Since in their case, the lengths have same values, this will be:

Perimeter = 2(l + l)

Perimeter = 4l

4l = 68

L = 68/4

L = 17m

Therefore, the dimensions are 17m and 17m.

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Step-by-step explanation:

The perimeter of a rectangle is expressed as 2(L+B) where;

L is the length and B is the breadth of the triangle.

P = 2(L+B)

68 = 2(L+B)

L+B = 68/2

L+B = 34

L = 34 - B ... 1

Area of the rectangle A = LB... 2

Substituting equation 1 into 2 will give;

A = (34-B)B

A = 34B-B²

To maximize the area of the triangle, dA/dB must be equal to zero i.e

dA/dB = 0

dA/dB = 34 - 2B = 0

34-2B = 0

2B = 34

Dividing both sides of the equation by 2 we will have;

B = 34/2

B = 17

Substituting B = 17 into equation 1 to get the length L

L = 34-17

L = 17m

This shows that the rectangle with maximum area is a square since L = B = 17m

The dimension of the rectangle is Length = 17m and Breadth = 17m

30

step-by-step explanation:

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