Answer :
(Refer to picture for diagram.)
Area of base [tex]=w^{2}[/tex]
Area of one side [tex]=\frac{8}{w^{2}}*w=\frac{8}{w}[/tex]
Cost of base material [tex]=8w^{2}[/tex]
Cost of material for a side [tex]=6*\frac{8}{w}=\frac{48}{w}[/tex]
There are 4 sides so [tex]4*\frac{48}{w}=\frac{192}{w}[/tex]
Total cost [tex]C(w)=8w^{2}+\frac{192}{w}[/tex]
Area of base [tex]=w^{2}[/tex]
Area of one side [tex]=\frac{8}{w^{2}}*w=\frac{8}{w}[/tex]
Cost of base material [tex]=8w^{2}[/tex]
Cost of material for a side [tex]=6*\frac{8}{w}=\frac{48}{w}[/tex]
There are 4 sides so [tex]4*\frac{48}{w}=\frac{192}{w}[/tex]
Total cost [tex]C(w)=8w^{2}+\frac{192}{w}[/tex]
The volume of a box is the amount of space in it.
The cost function is: [tex]\mathbf{C(w)=8w^2 + \frac{192}{w}}[/tex]
The volume of the box is:
[tex]\mathbf{V = 8}[/tex]
Assume the base length is w.
So, the area of the base is:
[tex]\mathbf{A_1 = w \times w}[/tex]
[tex]\mathbf{A_1 = w^2}[/tex]
The base material costs $8 per square meter.
So, the cost of the base material would be:
[tex]\mathbf{C_1 = 8w^2}[/tex]
Recall that:
[tex]\mathbf{V = 8}[/tex]
The volume of a box with a square base is:
[tex]\mathbf{V = w^2h}[/tex]
Make h the subject
[tex]\mathbf{h = \frac{V}{w^2}}[/tex]
Substitute 8 for V
[tex]\mathbf{h = \frac{8}{w^2}}[/tex]
The side area is:
[tex]\mathbf{A =4wh}[/tex]
So, we have:
[tex]\mathbf{A = 4 \times w \times \frac{8}{w^2}}[/tex]
[tex]\mathbf{A = \frac{32}{w}}[/tex]
The side materials cost $6 per square meter.
So, the cost of the side materials would be:
[tex]\mathbf{C_2 = 6 \times \frac{32}{w}}[/tex]
[tex]\mathbf{C_2 = \frac{192}{w}}[/tex]
The cost function of the box is:
[tex]\mathbf{C =C_1 + C_2}[/tex]
So, we have:
[tex]\mathbf{C=8w^2 + \frac{192}{w}}[/tex]
Hence, the cost function is:
[tex]\mathbf{C(w)=8w^2 + \frac{192}{w}}[/tex]
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