Answer:
The dimensions of the box is 3 ft by 3 ft by 30.22 ft.
The length of one side of the base of the given box is 3 ft.
The height of the box is 30.22 ft.
Step-by-step explanation:
Given that, a rectangular box with volume of 272 cubic ft.
Assume height of the box be h and the length of one side of the square base of the box is x.
Area of the base is = ![(x\times x)](https://tex.z-dn.net/?f=%28x%5Ctimes%20x%29)
![=x^2](https://tex.z-dn.net/?f=%3Dx%5E2)
The volume of the box is = area of the base × height
![=x^2h](https://tex.z-dn.net/?f=%3Dx%5E2h)
Therefore,
![x^2h=272](https://tex.z-dn.net/?f=x%5E2h%3D272)
![\Rightarrow h=\frac{272}{x^2}](https://tex.z-dn.net/?f=%5CRightarrow%20h%3D%5Cfrac%7B272%7D%7Bx%5E2%7D)
The cost per square foot for bottom is 20 cent.
The cost to construct of the bottom of the box is
=area of the bottom ×20
cents
The cost per square foot for top is 10 cent.
The cost to construct of the top of the box is
=area of the top ×10
cents
The cost per square foot for side is 1.5 cent.
The cost to construct of the sides of the box is
=area of the side ×1.5
cents
cents
Total cost = ![(20x^2+10x^2+6xh)](https://tex.z-dn.net/?f=%2820x%5E2%2B10x%5E2%2B6xh%29)
![=30x^2+6xh](https://tex.z-dn.net/?f=%3D30x%5E2%2B6xh)
Let
C![=30x^2+6xh](https://tex.z-dn.net/?f=%3D30x%5E2%2B6xh)
Putting the value of h
![C=30x^2+6x\times \frac{272}{x^2}](https://tex.z-dn.net/?f=C%3D30x%5E2%2B6x%5Ctimes%20%5Cfrac%7B272%7D%7Bx%5E2%7D)
![\Rightarrow C=30x^2+\frac{1632}{x}](https://tex.z-dn.net/?f=%5CRightarrow%20C%3D30x%5E2%2B%5Cfrac%7B1632%7D%7Bx%7D)
Differentiating with respect to x
![C'=60x-\frac{1632}{x^2}](https://tex.z-dn.net/?f=C%27%3D60x-%5Cfrac%7B1632%7D%7Bx%5E2%7D)
Again differentiating with respect to x
![C''=60+\frac{3264}{x^3}](https://tex.z-dn.net/?f=C%27%27%3D60%2B%5Cfrac%7B3264%7D%7Bx%5E3%7D)
Now set C'=0
![60x-\frac{1632}{x^2}=0](https://tex.z-dn.net/?f=60x-%5Cfrac%7B1632%7D%7Bx%5E2%7D%3D0)
![\Rightarrow 60x=\frac{1632}{x^2}](https://tex.z-dn.net/?f=%5CRightarrow%2060x%3D%5Cfrac%7B1632%7D%7Bx%5E2%7D)
![\Rightarrow x^3=\frac{1632}{60}](https://tex.z-dn.net/?f=%5CRightarrow%20x%5E3%3D%5Cfrac%7B1632%7D%7B60%7D)
![\Rightarrow x\approx 3](https://tex.z-dn.net/?f=%5CRightarrow%20x%5Capprox%203)
Now ![C''|_{x=3}=60+\frac{3264}{3^3}>0](https://tex.z-dn.net/?f=C%27%27%7C_%7Bx%3D3%7D%3D60%2B%5Cfrac%7B3264%7D%7B3%5E3%7D%3E0)
Since at x=3 , C''>0. So at x=3, C has a minimum value.
The length of one side of the base of the box is 3 ft.
The height of the box is ![=\frac{272}{3^2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B272%7D%7B3%5E2%7D)
=30.22 ft.
The dimensions of the box is 3 ft by 3 ft by 30.22 ft.