Question

6. A closed box with a square base is to contain 252 cubic feet. The bottom costs RM5 per square foot, the top costs RM2
per square foot, and the sides costs RM3 per square foot. Find the dimensions that will minimize the cost?
[width-base-6, height=7]
​

Answer 1

Answer:

The dimensions of the box is 6feet x 6feet x 7feet

Step-by-step explanation:

Let the length = breadth = l

and height = h

We know that for cuboid, Volume = Length * Breadth * Height

                                                  252 = l * l * h

                                                  252 = l²h

                                                      h = 252/ l²               --(i)

Area required = Top + Bottom + 4*Walls

                       = ( l * l ) + ( l * l ) + 4 * ( l * h )

                       = l² + l² + 4lh

           A         = 2l² + 4lh

According to the question, the cost = 2*Top + 5*Bottom + 3*(4*Walls)

                                                           = 2*( l * l ) + 5*( l * l ) + 3*[ 4 * ( l * h )]

                                                           = 2l² + 5l² + 12lh

                                                           = 7l² + 12lh

                                                           = 7l² + 12*l*(252/ l²)

                                                   C      = 7l² + 12*(252/ l)

$\frac{dC}{dl}$ = 14l + 3024 * ($\frac{-1}{l^{2}}$)

Taking $\frac{dC}{dl}$ = 0

=> 14l + 3024 * ($\frac{-1}{l^{2}}$) = 0

=> 14l = $\frac{3024}{l^{2}}$

=> l³ = 216

=> l = 6

$\frac{d^{2}C}{dl^{2}}$ = 14 -3024 ($\frac{-2}{l^{3}}$)

$\frac{d^{2}C}{dl^{2}}$ = 14 -3024 ($\frac{-2}{216}$)

$\frac{d^{2}C}{dl^{2}}$ = 14 +28

$\frac{d^{2}C}{dl^{2}}$ = 42 > 0             (Minima)

As,  h = 252/ l² and l = 6

=> h = 252/36 = 7

Therefore the dimensions of the box is 6feet x 6feet x 7feet