I would like to know how to solve the system coming from the Lagrange method to find maximum and minimum values. The problem is:

*Find the maximum and minimum volume for a rectangular closed box with surface area $\,1500\,\, cm^2\,$ and sum of edges length $\,200\,\, cm$.

So, here are my functions: $V(x,y,z)=xyz$, $A(x,y,z)=2(xy+xz+yz)$ and $L(x,y,z)=4(x+y+z)$. The system is \begin{align}yz&=2\lambda(y+z)+4\mu\\ xz&=2\lambda(x+z)+4\mu\\ xy&=2\lambda(x+y)+4\mu\\ 1500&=2(xy+xz+yz)\\200&=4(x+y+z)\end{align}

Well, what is the best approach to solve? Thanks.


There is a solution which can be obtained using Lagrange multipliers, but we need to take care of some nuances when obtaining a solution.

We need to check that the constraint set is non-empty. The length constraint can be used to eliminate $z$ using $z = 50-(x+y)$. If we set $y=11$, the area constraint becomes (after eliminating $z$): $$ -2 x^2+78x = 642.$$ Substituting $x=11$ and $x=12$ and using the intermediate value theorem shows that the constraint set is non-empty.

We can observe that the length constraint can be manipulated to get $$ (x+y+z)^2 = 2,500 = x^2+y^2+z^2 + 2 ( xy + yz+ zx) = x^2+y^2+z^2 + 1,500,$$ which gives $x^2+y^2+z^2 = 1,000$. Thus the feasible set is bounded, and trivially closed, hence compact. This means that the volume has a minimum and a maximum.

Some other observations about the solutions: First, a quick check will show that there is no solution with all components equal that satisfies both constraints. Second, it is straightforward to check that there is no solution that satisfies the constraints if any of the components are zero. Third, it is straightforward to check that the constraint gradients are linearly dependent iff all components are equal. Thus the first observation ensures that the constraint gradients are linearly independent and we can use the multiplier method. Fourth, notice that the cost and constraints have the same value if the components are permuted.

In the context of this problem, the Lagrange multipliers give qualitative rather than quantitative information about the solution.

Now observe that the Lagrange equations can be written as $yz-2\lambda(y+z)=4\mu$. Thus we can take two equations and get $yz-2\lambda(y+z)= x z - 2 \lambda ( x+z)$. Rearranging gives $z ( y-x) - 2 \lambda ( y-x) = 0$, from which we get the following equation: $$z ( y-x) = 2 \lambda ( y-x).$$ Repeating the process for the other two pairs yields similar equations.

Now note that if all $x,y,z$ are distinct, then the above equation(s) result in $x = y = z = 2 \lambda$, which is an immediate contradiction. We have already observed that there is no solution with all components equal, from which we conclude that at an extremum, two components are equal, and the third different. From symmetry, we may take $z = y$, which simplifies the equations to: \begin{align} x+2y &= 50 \\ 2 xy+y^2 &= 750. \end{align} Using the first equation, we obtain a quadratic equation in $y$: $$ -3 y^2+100y=750,$$ which yields the solutions $y = \frac{50 \pm 5 \sqrt{10}}{3}$. Substituting for $x$ gives $x = \frac{50 \mp 10 \sqrt{10}}{3}$. Using $z=y$ gives the corresponding volumes as $V = \frac{87,500 \pm 2,500 \sqrt{10}}{3^3}$ (it is straightforward to check that both constraints are satisfied). Since we know that $V$ has a minimum and a maximum at which the Lagrange conditions must be satisfied, then these are the corresponding extreme values for volume.

For those who care, it is straightforward to compute the corresponding multiplier values; $\lambda = \frac{50 \mp 5 \sqrt{10}}{6}, \mu = \frac{-1375 \pm 250 \sqrt{10}}{18}$.

  • $\begingroup$ I couldn't imagine how many facts to observe during the solution. Thanks so much for your effort. Why did you look to $y=11$? $\endgroup$ – Sigur Jun 24 '12 at 15:15

Rewrite the first equation as $yz-2\lambda y -2\lambda z=4\mu$, and then as $(y-2\lambda)(z-2\lambda)-4\lambda^2=4\mu$, and then as $(y-2\lambda)(z-2\lambda)=4\lambda^2+4\mu$.

Do the same for the next two equations. We conclude that $$(y-2\lambda)(z-2\lambda)=(x-2\lambda)(z-2\lambda)=(x-2\lambda)(y-2\lambda).\tag{$1$}$$

The equations $(1)$ are very useful for solving the system. One needs to be a little careful. Note that $ab=ac$ if and only if $b=c$ or $a=0$.

  • $\begingroup$ Thanks. The first step is the crucial one. Now, everything follows. Bye. $\endgroup$ – Sigur Jun 23 '12 at 16:14

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