Question

The base of an aquarium with given volume\(V\)is made of slate and the sides are made of glass. If slate costs five times as much (per unit area) as glass, find the dimensions of the aquarium that minimize the cost of the materials.

Short answer

\(x = {\left( {\frac{{2v}}{5}} \right)^{1/3}},\;\;\;y = {\left( {\frac{{2v}}{5}} \right)^{1/3}},z = \frac{5}{2}{\left( {\frac{{2v}}{5}} \right)^{1/3}}\)

Step-by-step solution

Second derivative test

Suppose the second partial derivatives of\(f\)are continuous on a disk with center\({\bf{(a,b)}}\), and suppose that\({f_x}(a,b) = 0\)and\({f_y}(a,b) = 0\). Let

\(D = D(a,b) = {f_{xx}}(a,b){f_{yy}}(a,b) - {\left( {{f_{xy}}(a,b)} \right)^2}\)

(a) If\(D > 0\)and\({f_{xx}}(a,b) > 0\), then\(f(a,b)\)is a local minimum.

(b) If\(D > 0\)and\({f_{xx}}(a,b) < 0\), then\(f(a,b)\)is a local maximum.

(c) If\(D < 0\), then\(f(a,b)\)is not a local maximum or minimum.

Compute volume

Let the dimensions of the base of the aquarium are\(x,y\)and its height be\(z\).

Then the volume\(v = xyz \cdots (1)\)

Let the cost of making the walls is 1 per unit area then the cost of making the base is 5 per unit area.

So, the total cost of making aquarium is\(c = 5xy + 2xz + 2yz\)

But\(z = \frac{v}{{xy}}\,\,(from(1))\)then

\(\begin{array}{c}c = 5xy + 2x\frac{v}{{xy}} + 2y\frac{v}{{xy}}\\ = 5xy + \frac{{2v}}{y} + \frac{{2v}}{x}\end{array}\)

Compute partial derivative

Take\(f(x,y) = 5xy + \frac{{2v}}{y} + \frac{{2v}}{x}\)

Then \({f_x} = 5y - \frac{{2v}}{{{x^2}}}\)

\({f_y} = 5x - \frac{{2v}}{{{y^2}}}\)

Find critical point

For critical point put\({f_x} = 0,\;\;\;{f_y} = 0\).

i.e\(5y - \frac{{2v}}{{{x^2}}} = 0\)

And\(5x - \frac{{2v}}{{{y^2}}} = 0\)

On comparing we have\(x = y\).

\(5{x^3} = 2v\)

i.e. \(\begin{array}{l}{x^3} = \frac{{2v}}{5}\\x = {\left( {\frac{{2v}}{5}} \right)^{1/3}}\end{array}\)

Also\(y = {\left( {\frac{{2v}}{5}} \right)^{1/3}}\)

Then the critical point is\(\left( {{{\left( {\frac{{2v}}{5}} \right)}^{1/3}},{{\left( {\frac{{2v}}{5}} \right)}^{1/3}}} \right)\).

Compute discriminant

Now \({f_{xx}} = \frac{{4v}}{{{x^3}}},\;\;\;{f_{yy}} = \frac{{4v}}{{{y^3}}},\;\;\;{f_{xy}} = 5\)

Then,

\(\begin{array}{c}D = {f_{xx}}{f_{yy}} - f_{xy}^2\\ = \frac{{16{v^2}}}{{{x^3}{y^3}}} - 25\end{array}\)

Compute discriminant at critical point

\({\rm{At }}\left( {{{\left( {\frac{{2v}}{5}} \right)}^{1/3}},{{\left( {\frac{{2v}}{5}} \right)}^{1/3}}} \right)\)

\(\begin{array}{c}D = \frac{{16{v^2}}}{{4{v^2}}} \times 25 - 25\\ = 75 > 0\end{array}\)

And\({f_{xx}} = 10 > 0\)

So, \(\left( {{{\left( {\frac{{2v}}{5}} \right)}^{1/3}},{{\left( {\frac{{2v}}{5}} \right)}^{1/3}}} \right)\) is a point of minima.

Dimension of aquarium

When \(x = {\left( {\frac{{2v}}{5}} \right)^{1/3}},y = {\left( {\frac{{2v}}{5}} \right)^{1/3}}\)the cost of the aquarium will be minimum.

Hence the dimensions of the aquarium for the cost to be minimum are

\(x = {\left( {\frac{{2v}}{5}} \right)^{1/3}},\;\;\;y = {\left( {\frac{{2v}}{5}} \right)^{1/3}},z = \frac{5}{2}{\left( {\frac{{2v}}{5}} \right)^{1/3}}\)