Question

A rectangular box with a volume of 684 ftcubed is to be constructed with a square base and top. The cost per square foot for the bottom is 20cents​, for the top is 15cents​, and for the sides is 1.5cents. What dimensions will minimize the​ cost?

Answer 1

Answer:

The dimensions of the rectangular box is 36.23 ft×36.23 ft×4.345 ft.  

Minimum cost=2046.16 cents      

Step-by-step explanation:

Given that a rectangular box with a volume of 684 ft³.

The base and the top of the rectangular box is square in shape

Let the length and width of the rectangular box be x.

[since the base is square in shape,  length=width]

and the height of the rectangular box be h.

The volume of rectangular box is = Length ×width × height

                                                      =(x²h) ft³

$x^2h=684\Rightarrow h=\frac{684}{x^2}$  (1)

The area of the base and top of rectangular box is = x² ft²

The surface area of the sides= 2(length+width) height

                                               =2(x+x)h

                                               =4xh ft²

The total cost to construct the rectangular box is

=[(x²×20)+(x²×15)+(4xh×1.5)] cents  

=(20x²+15x²+6xh) cents

=(25x²+6xh) cents

Total cost= C(x).

C(x) is in cents.

∴C(x)=25x²+6xh

Putting $h=\frac{684}{x^2}$

$C(x)=25x^2+6x\times\frac{684}{x^2} \Rightarrow C(x)=25x^2+\frac{4104}{x}$

Differentiating with respect to x

$C'(x)=50x-\frac{4104}{x^2}$

To find minimum cost, we set C'(x)=0

$\therefore50x-\frac{4104}{x^2}=0\\\Rightarrow50x=\frac{4104}{x^2}\\\Rightarrow x^3=\frac{4104}{50}\Rightarrow x\approx 4.345$ ft.

Putting the value x in equation (1) we get

$h=\frac{684}{(4.345)^2}$

 ≈36.23 ft.

The dimensions of the rectangular box is 36.23 ft×36.23 ft×4.345 ft.  

Minimum cost C(x)=[25(4.345)²+10(4.345)(36.23)] cents

                               =2046.16 cents