Jtdylktuy

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:

$begin{align*} nabla A=frac{1}{2}langle 12-2x,,13-2yrangle &= lambdalangle4x-24,, -4y+26rangle\ implies&begin{cases} -x+6=lambda(4x-24)\-y+frac{13}{2}=lambda(-4y+26)\x^2+(12-x)^2-y^2-(13-y)^2=0end{cases}end{align*}$

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.

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 $4x-24neq 0$ and when $-4y+26neq 0$. 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:
$$6^2+(12-6)^2-y^2-(13-y)^2=0rightarrow y=frac{13}{2}pm frac{5i}{2}$$
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:
$$x^2+(12-x)^2-left(frac{13}2right)^2-left(13-left(frac{13}2right)right)^2=0rightarrow x=frac 7 2 text{ or } x=frac{17}2$$

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.

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:
$$
L(x,y,lambda)=frac{1}{2} ((12-x) x+(13-y) y)-lambda
left(x^2+(12-x)^2-y^2-(13-y)^2right)
$$
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:
$$
nabla^2f=left(begin{array}{cc}-1& 0 \ 0 & -1end{array}right)
$$

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

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

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