Function Optimization with Non-linear Constraint, Lagrange Multipliers Fails

Question

I am trying to maximize the function $A(x,y)=\frac{1}{2}(x(12-x)+y(13-y))$ subject to the constraint $x^2+(12-x)^2-y^2-(13-y)^2=0$.

My attempt:

$\begin{align*} \nabla A=\frac{1}{2}\langle 12-2x,\,13-2y\rangle &= \lambda\langle4x-24,\, -4y+26\rangle\\ \implies&\begin{cases} -x+6=\lambda(4x-24)\\-y+\frac{13}{2}=\lambda(-4y+26)\\x^2+(12-x)^2-y^2-(13-y)^2=0\end{cases}\end{align*}$

But clearly there is no solution due to the first two equations.

Using Wolfram Alpha, however, yields a maximum at $\displaystyle \left(\frac{17}{2},\,\frac{13}{2}\right)$ being $A=36$ and shows a nice little graph.

Answer 1

First, let's re-explain why the first two equations are contradictory. The first equation gives us:

$$\lambda=\frac{-x+6}{4x-24}=-\frac 1 4$$

While the second equation gives us:

$$\lambda=\frac{-y+\frac{13} 2}{-4y+26}=\frac 1 4$$

However, what's important to realize here is that these two equations only work when $4x-24\neq 0$ and when $-4y+26\neq 0$. Therefore, in order to find a solution, we need to consider the other cases: $4x-24=0$ (i.e. $x=6$) and $-4y+26=0$ (i.e. $y=\frac {13} 2$).

Case 1: $x=6$

Let's plug $x=6$ into our constraint equation: $$6^2+(12-6)^2-y^2-(13-y)^2=0\rightarrow y=\frac{13}{2}\pm \frac{5i}{2}$$ Thus, this equation has no real solutions, and this case can be ignored.

Case 1: $y=\frac {13} 2$

Let's plug $y=\frac {13} 2$ into our constraint equation: $$x^2+(12-x)^2-\left(\frac{13}2\right)^2-\left(13-\left(\frac{13}2\right)\right)^2=0\rightarrow x=\frac 7 2 \text{ or } x=\frac{17}2$$

Thus, we have two critical points: $(\frac 7 2, \frac{13}2)$ and $(\frac{17}2, \frac{13}2)$. I will leave it to you to show these critical points are maximums.

Answer 2

For a constrained problem \begin{align} \max{} & f(x,y) \\ & g(x,y)\le 0 \end{align} you must write the Lagrangian: $$ L(x,y,\lambda)=f(x,y)+\lambda g(x,y) $$ Then you must find stationary points of your Lagrangian (attention this is only necessary conditions, see wiki) \begin{align} \partial_x L &= 0 = \partial_x f+ \lambda \partial_x g \\ \partial_y L &= 0 = \partial_y f+ \lambda \partial_y g \\ \partial_\lambda L &= 0 = g \end{align} With your example, this is essentially computations: $$ L(x,y,\lambda)=\frac{1}{2} ((12-x) x+(13-y) y)-\lambda \left(x^2+(12-x)^2-y^2-(13-y)^2\right) $$ your three equations are (after simplification): \begin{align} -(-6 + x) (1 + 4 \lambda) &= 0 \\ \frac{1}{2} (-13 + 2 y) (-1 + 4 \lambda) &= 0\\ -x^2-(12-x)^2+y^2+(13-y)^2 &=0 &\\ \end{align} The (real) solutions are: $$ (x,y,\lambda)=(\frac{7}{2},\frac{13}{2},-\frac{1}{4}) $$ and $$ (x,y,\lambda)=(\frac{17}{2},\frac{13}{2},-\frac{1}{4}) $$ You can check that for these two solutions $$ f(\frac{7}{2},\frac{13}{2})=f(\frac{17}{2},\frac{13}{2})=36 $$

Extra: to check that these points are maximizers you must check that the Hessian of $f(x,y)$ is symmetric definite negative, which is clearly the case as: $$ \nabla^2f=\left(\begin{array}{cc}-1& 0 \\ 0 & -1\end{array}\right) $$

Answer 3

Let $x(12-x)=a$ and $y(13-y)=b$.

Thus, the condition gives $b=a+12.5$.

Also, we have $$a=x(12-x)\leq\left(\frac{x+12-x}{2}\right)=36$$ and $$b=y(13-y)\leq\left(\frac{y+13-y}{2}\right)=42.25,$$ which gives $$a=b-12.5\leq42.25-12.5=29.75.$$ Id est, $$A(x,y)=\frac{1}{2}(a+b)=a+6.25\leq29.75+6.25=36.$$ The equality occurs for $b=42.25$ or $y=6.5$, which says that we got a maximal value.