Answer:
Length = Width = 7 inches
Height = 14 inches
Step-by-step explanation:
Let's call
x = length and width of the bottom and top
y = heigth
The bottom and top are squares, so their total area is
[tex]A_{bt}=x^2+x^2=2x^2[/tex]
the area of one side is xy. As we have 4 sides, the total area of the sides is
[tex]A_s=4xy[/tex]
The cost for the bottom and top would be
[tex]C_{bt}=\$4.2x^2=\$8x^2[/tex]
The cost for the sides would be
[tex]C_s=\$2.4xy=\$8xy[/tex]
The total cost to made the box is
[tex]C_t=\$(8x^2+8xy)[/tex]
But the volume of the box is
[tex]V=x^2y=686in^3[/tex]
isolating
[tex]y=\frac{686}{x^2}[/tex]
Replacing this expression in the formula of the total cost
[tex]C_t(x)=8x^2+8x(\frac{686}{x^2})=8x^2+\frac{5488}{x}[/tex]
The minimum of the cost should be attained at the point x were the derivative of the cost is zero.
Taking the derivative
[tex]C'_t(x)=16x-\frac{5488}{x^2}[/tex]
If C' = 0 then
[tex]16x=\frac{5488}{x^2}\Rightarrow x^3=\frac{5488}{16}\Rightarrow x^3=343\Rightarrow x=\sqrt[3]{343}=7[/tex]
We have to check that this is actually a minimum.
To check this out, we take the second derivative
[tex]C''_t(x)=16+\frac{10976}{x^3}[/tex]
Evaluating this expression in x=7, we get C''>0, so x it is a minimum.
Now that we have x=7, we replace it on the equation of the volume to get
[tex]y=\frac{686}{49}=14in[/tex]
The dimensions of the most economical box are
Length = Width = 7 inches
Height = 14 inches
