Question

The base of an aquarium with given volume V is made of slate and the sides are made of glass. If the slate costs seven times as much (per unit area) as glass, use Lagrange multipliers to find the dimensions of the aquarium that minimize the cost of the materials. (Enter the dimensions as a comma separated list; Note that the variable volume is a capital V and must be entered as such, using the shift rather than caps-lock key)

Answer 1

Answer:

$l = \sqrt[3]{\frac{2V}{7}}$     $b = \sqrt[3]{\frac{2V}{7}}$       $h = \sqrt[3]{\frac{49V}{4}}$

Step-by-step explanation:

Represent the volume of the box with V and the dimensions with l, b and h.

The volume (V) is:

$V = l * b * h$

Make h the subject of the formula

$h = \frac{V}{lb}$

The surface area (S) of the aquarium is:

$S = lb + 2(lh + bh)$

Where lb represents the area of the base (i.e. slate):

The cost (C) of the surface area is:

$C = 7 * lb + 1 * 2(lh + bh)$

$C = 7lb + 2(lh + bh)$

$C = 7lb + 2h(l + b)$

Substitute $\frac{V}{lb}$ for h in the above equation

$C = 7lb + 2*\frac{V}{lb}(l + b)$

$C = 7lb + \frac{2V}{lb}(l + b)$

$C = 7lb + \frac{2V}{b} + \frac{2V}{l}$

Differentiate with respect to l and with respect to b

$C_l=7b - \frac{2V}{l^2}$ $=0$

$C_b=7l - \frac{2V}{b^2}$ $=0$

To solve for b and l, we equate both equations and set l to b (to minimize the cost)

$7b - \frac{2V}{l^2}=7l - \frac{2V}{b^2}$

$7l - \frac{2V}{l^2}=7b - \frac{2V}{b^2}$

By comparison:

$l =b$

$C_l=7b - \frac{2V}{l^2}$ $=0$ becomes

$7l - \frac{2V}{l^2}=0$

$7l = \frac{2V}{l^2}$

Cross Multiply

$7l^3 = 2V$

Solve for l

$l^3 = \frac{2V}{7}$

$l = \sqrt[3]{\frac{2V}{7}}$

Recall that: $l =b$

$b = \sqrt[3]{\frac{2V}{7}}$

Also recall that:

$h = \frac{V}{lb}$

$h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}$

$h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}$

Apply law of indices

$h = \sqrt[3]{\frac{49V^3}{4V^2}}$

$h = \sqrt[3]{\frac{49V}{4}}$

The dimension that minimizes the cost of material of the aquarium is:

$l = \sqrt[3]{\frac{2V}{7}}$     $b = \sqrt[3]{\frac{2V}{7}}$       $h = \sqrt[3]{\frac{49V}{4}}$