Question

A rectangular frame is to be made of 240 cm long. Determine the value of the length of the rectangle required to maximize the area.

(a) 24 cm

(b) 60 cm

(c) 240 cm

(d) 120 cm

This question was posed to me in an online interview.

I'd like to ask this question from First Order Nonlinear Differential Equations topic in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right answer is (b) 60 cm

Easy explanation: Let us consider ‘x’ as length and ‘y’ as the breadth of the rectangle.

Given: Perimeter 2(x + y) = 240 cm

       x + y = 120

       y = 120 – x

Area of the rectangle, a = x*y = x(120-x) = 120x – x^2

Finding the derivative, we get, \(\frac{d(a)}{dx}= \frac{d(120x- x^2)}{dx}=120-2x \)

To find the value of x that maximizes the area, we substitute \(\frac{d(a)}{dx}= 0.\)

Therefore, we get, 120 – 2x =0

       2x = 120

    x = 60 cm

To check if x = 60 cm is the value that maximizes the area, we find the second derivative of the area,

\(\frac{d^2 (a)}{dx^2}= -2\)  < 0 …………………. (i)

We know that the condition for maxima is \(\frac{d^2 (f(x))}{dx^2}<0,\) which is satisfied by (i), therefore, x = 60 cm maximizes the area of the rectangle.