Question

Find the maximum and minimum volumes of a rectangular box whose surface area is \(150\mathop {cm}\nolimits^2 \) and whose total edge length is\(200cm\).

Short answer

Maximum volume is \(3534{\bf{c}}{{\bf{m}}^3}\), Minimum volume is \(2948{\bf{c}}{{\bf{m}}^3}\)

Step-by-step solution

Step-1

Let the three non-equal edges be \({\rm{a, b, c}}\)

Surface areais

\(p(a,b,c) = 2ab + 2bc + 2ca = 1500\;{\rm{c}}{{\rm{m}}^2}\) ( Given )

\({\rm{p(a, b, c) = a b + b c + c a = 750}}\)

The total edge length is \({\rm{q(a, b, c) = 4 a + 4 b + 4 c = 200}}\)

\({\rm{q(a, b, c) = a + b + c = 50}}\)

We need to find the Minimumand Maximumvalue \({\rm{V(a, b, c) = a b c}}\)which is the volume

Step-2

To find the maximum/minimum volume, we will use Lagrange's multiplier method:

\(\begin{array}{l}{V_a} = bc = \lambda {p_a} + \rho \times {q_a} = \lambda (b + c) + \rho \\{V_b} = ac = \lambda {p_b} + \rho \times {q_b} = \lambda (a + c) + \rho \\{V_c} = ab = \lambda {p_c} + \rho \times {q_c} = \lambda (a + b) + \rho \end{array}\)

This gives us:

\(\begin{array}{l}\lambda b + \lambda c + \rho = bc \to (1)\\\lambda a + \lambda c + \rho = ac \to (2)\\\lambda a + \lambda b + \rho = ab \to (3)\end{array}\)

Step-3

Subtract equation (1) from equation(2), To get

\(\lambda (b - a) = c(b - a) \to (4)\)

Subtract equation (2) from equation(3), To get

\(\lambda (c - b) = a(c - b) \to (5)\)

Subtract equation (1) from equation(3), To get

\(\lambda (c - a) = b(c - a) \to (6)\)

From Eqn(4), Eqn(5), and Eqn(6), We have \(\lambda = a = b\) OR \(\lambda = a = c{\rm{OR}}\lambda = c = b\)

When we divide both sides by \({\rm{b - a}}\) in Eqn(4), we get

\(\lambda = c \to \left( {{4^\prime }} \right)\)

And when we divide both sides by \({\rm{c - b}}\) in Eqn(5), we get

\(\lambda = a \to \left( {{5^\prime }} \right)\)

But now we CAN NOT! Divide both sides by \({\rm{c - a}}\)in Eqn \({\rm{6}}\) because by Eqn \(\left( {{4^\prime }} \right)\)and Eqn \(\left( {{5^\prime }} \right)\) we have \({\rm{c = a}}\)and hence \({\rm{c - a = 0}}\)

Note that there is nothing special about \({\rm{a}}\)which is different from \({\rm{b}}\) or\({\rm{c}}\). Therefore by symmetry, we got the three cases separated by 'OR'

Step-4

We are taking \(\lambda = a = b\)

\(\begin{array}{l}p(a,b,c) = ab + bc + ca = 750\\p(a,a,c) = {a^2} + 2ac = 750 \to (7)\\q(a,b,c) = a + b + c = 50\\q(a,a,c) = 2a + c = 50 \to {\rm{ (8) }}\end{array}\)

Using equations (7) and (8), we have

\(\begin{array}{l}{a^2} + 2a(50 - 2a) = 750\\100a - 3{a^2} = 750\end{array}\)

Using the quadratic equation, we have \(a = \frac{{5(10 \pm \sqrt {10} )}}{3}\)

Volume when \(a = \frac{{5(10 + \sqrt {10} )}}{3}\) is

\(\begin{array}{l}abc = {a^2}c = {a^2}(50 - 2a) = \\{\left( {\frac{{5(10 + \sqrt {10} )}}{3}} \right)^2}\left( {50 - \frac{{10(10 + \sqrt {10} )}}{3}} \right) \approx 2948\end{array}\)

Centimeter cube Volume when \(a = \frac{{5(10 - \sqrt {10} )}}{3}\) is

\({\left( {\frac{{5(10 - \sqrt {10} )}}{3}} \right)^2}\left( {50 - \frac{{10(10 - \sqrt {10} )}}{3}} \right) \approx 3534\)Centimeter cube

Therefore, the Maximum volume is \(3534{\bf{c}}{{\bf{m}}^3}\), Minimum volume is \(2948{\bf{c}}{{\bf{m}}^3}\)