Question

The length and width of a rectangle must have a sum of 60. Find the dimensions of the rectangle that will have the maximum area. [Hint: Let x and 60-x be the length
and width. The area can be described by the function f(x)=x(60-x).]
The length is… and the width is…

Answer 1

If the sum of the length and width of rectangle is 60 and rectangle is having maximum area then the dimensions are 30 units each.

Given that the sum of length and breadth of rectangle is 60.

We are required to find the dimensions of the rectangle that will have the maximum area. Area is basically how much part of surface is being covered by that particular shape or substance.

Let the length of rectangle be x.

According to question the breadth will be (60-x).----2

Area of rectangle=Length *Breadth

A=x(60-x)

A=60x-$x^{2}$

Differentiate A with respect to x.

dA/dx=60-2x

Again differentiate with respect to x.

$d^{2} A/dA^{2}$=-2x

-2x<0

So the area is maximum because x cannot be less than or equal to 0.

Put dA/dx=0

60-2x=0

60=2x

x=30

Put the value of x in 2 to get the breadth.

Breadth=60-x

=60-30

=30

Hence if the sum of the length and width of rectangle is 60 and rectangle is having maximum area then the dimensions are 30 units each.

Learn more about differentiation at https://brainly.com/question/954654

#SPJ1