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
[tex]x^2h=684\Rightarrow h=\frac{684}{x^2}[/tex] (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 [tex]h=\frac{684}{x^2}[/tex]
[tex]C(x)=25x^2+6x\times\frac{684}{x^2} \Rightarrow C(x)=25x^2+\frac{4104}{x}[/tex]
Differentiating with respect to x
[tex]C'(x)=50x-\frac{4104}{x^2}[/tex]
To find minimum cost, we set C'(x)=0
[tex]\therefore50x-\frac{4104}{x^2}=0\\\Rightarrow50x=\frac{4104}{x^2}\\\Rightarrow x^3=\frac{4104}{50}\Rightarrow x\approx 4.345[/tex] ft.
Putting the value x in equation (1) we get
[tex]h=\frac{684}{(4.345)^2}[/tex]
≈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
