Question

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.)

Answer 1

Answer:

Length is 17m and Breadth is also 17m

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