Question

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off squares from each corners and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?

Answer 1

Length of rectangle sheet = 45 cm 

Breath of rectangle sheet = 24 cm 

Let the side length of each small square be a. 

If by cutting a square from each corner and folding up the flaps we will get a cuboidal box with 

Length, L = 45 – 2a 

Breadth, B = 24 – 2a and 

Height, H = a 

Assuming, 

Volume of box, 

V = LBH = (45 - 2a)(24 - 2a)(a) 

Condition for maxima and minima is,

\(\frac{dV}{da}\) = 0

⇒ (45 - 2a)(24 - 2a) + ( - 2)(24 - 2a)(a) + (45 - 2a)( - 2)(a) = 0 

⇒ 4a² – 138a + 1080 + 4a² – 48a + 4a² – 90a = 0 

⇒ 12a² – 276a + 1080= 0 

⇒ a² – 23a + 90 = 0 

⇒ a = 5, 18

\(\frac{d^2V}{da^2}\) = 24a - 276

For a = 5,

\(\frac{d^2V}{da^2}\) = -156,

⇒ \(\frac{d^2A}{d\theta^2}\) < 0

For a = 18,

 \(\frac{d^2V}{da^2}\) = +156,

⇒ \(\frac{d^2A}{d\theta^2}\) >0

So for A to maximum,

\(\frac{d^2A}{d\theta^2}\) < 0

Hence, 

a = 5 will give maximum volume. 

And maximum volume 

V = (45 - 2a)(24 - 2a)(a) 

= 2450 cm³