Respuesta :
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 = [tex](x\times x)[/tex]
[tex]=x^2[/tex]
The volume of the box is = area of the base × height
[tex]=x^2h[/tex]
Therefore,
[tex]x^2h=272[/tex]
[tex]\Rightarrow h=\frac{272}{x^2}[/tex]
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
[tex]=20x^2[/tex] 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
[tex]=10x^2[/tex] 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
[tex]=4xh\times 1.5[/tex] cents
[tex]=6xh[/tex] cents
Total cost = [tex](20x^2+10x^2+6xh)[/tex]
[tex]=30x^2+6xh[/tex]
Let
C[tex]=30x^2+6xh[/tex]
Putting the value of h
[tex]C=30x^2+6x\times \frac{272}{x^2}[/tex]
[tex]\Rightarrow C=30x^2+\frac{1632}{x}[/tex]
Differentiating with respect to x
[tex]C'=60x-\frac{1632}{x^2}[/tex]
Again differentiating with respect to x
[tex]C''=60+\frac{3264}{x^3}[/tex]
Now set C'=0
[tex]60x-\frac{1632}{x^2}=0[/tex]
[tex]\Rightarrow 60x=\frac{1632}{x^2}[/tex]
[tex]\Rightarrow x^3=\frac{1632}{60}[/tex]
[tex]\Rightarrow x\approx 3[/tex]
Now [tex]C''|_{x=3}=60+\frac{3264}{3^3}>0[/tex]
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 [tex]=\frac{272}{3^2}[/tex]
=30.22 ft.
The dimensions of the box is 3 ft by 3 ft by 30.22 ft.
