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?

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

 (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

Differentiating with respect to x

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

ft.

Putting the value x in equation (1) we get

 ≈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      

here the number -5 is all of the following except a rational number.

step-by-step explanation:

no, because the sides are not congruent