Question

What is the maximum area of the rectangle with perimeter 620 mm?

(a) 24,025 mm^2

(b) 22,725 mm^2

(c) 24,000 mm^2

(d) 24,075 mm^2

This question was posed to me by my school principal while I was bunking the class.

My enquiry is from Derivatives of Area in division Differential Calculus of Engineering Mathematics

Answer 1

Right answer is (a) 24,025 mm^2

Explanation: Let x be the length of the rectangle and y be the width of the rectangle. Then, Area A is,

A=x*y              …………………………………………………. (1)

Given: Perimeter of the rectangle is 620 mm. Therefore,

P=2(x+y)

620=2(x+y)

x+y=310

y=310-x

We can now substitute the value of y in (1)

A=x*(310-x)

A=310x-x^2

To find maximum value we need derivative of A,

dA/dx=310-2x

To find maximum value, \(\frac{dA}{dx}=0 \)

310-2x=0

2x=310

x=155 mm

Therefore, when the value of x=155 mm and the value of y=310-155=155 mm, the area of the rectangle is maximum, i.e., A=155*155=24,025 mm^2