Answer:
[tex]l = \sqrt[3]{\frac{2V}{7}}[/tex] [tex]b = \sqrt[3]{\frac{2V}{7}}[/tex] [tex]h = \sqrt[3]{\frac{49V}{4}}[/tex]
Step-by-step explanation:
Represent the volume of the box with V and the dimensions with l, b and h.
The volume (V) is:
[tex]V = l * b * h[/tex]
Make h the subject of the formula
[tex]h = \frac{V}{lb}[/tex]
The surface area (S) of the aquarium is:
[tex]S = lb + 2(lh + bh)[/tex]
Where lb represents the area of the base (i.e. slate):
The cost (C) of the surface area is:
[tex]C = 7 * lb + 1 * 2(lh + bh)[/tex]
[tex]C = 7lb + 2(lh + bh)[/tex]
[tex]C = 7lb + 2h(l + b)[/tex]
Substitute [tex]\frac{V}{lb}[/tex] for h in the above equation
[tex]C = 7lb + 2*\frac{V}{lb}(l + b)[/tex]
[tex]C = 7lb + \frac{2V}{lb}(l + b)[/tex]
[tex]C = 7lb + \frac{2V}{b} + \frac{2V}{l}[/tex]
Differentiate with respect to l and with respect to b
[tex]C_l=7b - \frac{2V}{l^2}[/tex] [tex]=0[/tex]
[tex]C_b=7l - \frac{2V}{b^2}[/tex] [tex]=0[/tex]
To solve for b and l, we equate both equations and set l to b (to minimize the cost)
[tex]7b - \frac{2V}{l^2}=7l - \frac{2V}{b^2}[/tex]
[tex]7l - \frac{2V}{l^2}=7b - \frac{2V}{b^2}[/tex]
By comparison:
[tex]l =b[/tex]
[tex]C_l=7b - \frac{2V}{l^2}[/tex] [tex]=0[/tex] becomes
[tex]7l - \frac{2V}{l^2}=0[/tex]
[tex]7l = \frac{2V}{l^2}[/tex]
Cross Multiply
[tex]7l^3 = 2V[/tex]
Solve for l
[tex]l^3 = \frac{2V}{7}[/tex]
[tex]l = \sqrt[3]{\frac{2V}{7}}[/tex]
Recall that: [tex]l =b[/tex]
[tex]b = \sqrt[3]{\frac{2V}{7}}[/tex]
Also recall that:
[tex]h = \frac{V}{lb}[/tex]
[tex]h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}[/tex]
[tex]h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}[/tex]
Apply law of indices
[tex]h = \sqrt[3]{\frac{49V^3}{4V^2}}[/tex]
[tex]h = \sqrt[3]{\frac{49V}{4}}[/tex]
The dimension that minimizes the cost of material of the aquarium is:
[tex]l = \sqrt[3]{\frac{2V}{7}}[/tex] [tex]b = \sqrt[3]{\frac{2V}{7}}[/tex] [tex]h = \sqrt[3]{\frac{49V}{4}}[/tex]
