Lagrange Maxima I am sorry to post this again, but I am still confused. 
Alice has only $24$ hours to study for an exam, and without preparation she will get $200$ points out of $1000$ points on the exam. It is estimated that her exam score will improve by $x(50−x)$ points if she reads her lecture notes for $x$ hours and $y(48−y)$ points if she solves review problems for y hours, but due to fatigue she will lose $(x+y)^2$ points. What is the maximum exam score she can obtain? Round your answer to the nearest integer.
I got $y=23/2$ and $x=25/2$.
When I plugged those back in I got $312.5$, and then I need to add the $200$ points. So my answer would be $512.5$ which would round to $513$. But this is not correct. 
I asked this question earlier and got $y=23/3$ and $x=26/3$ with it equaling $601$.
Can someone explain how they got these and if that is indeed correct?
 A: I know that this question is tagged as a Lagrange multiplier question, but as it turns out, you don't need a Lagrange multiplier because the inequality ($x+y\le 24$) is not tight, so it can be ignored.
The derivation is really simple. The total points are given by the equation,
$$p=200+x(50-x)+y(48-y)-(x+y)^2.$$
At the maximum, $p$ is extremal with respect to both $x$ and $y$, thus partial derivatives of $p$ with respect to both variables must be zero:
$$\frac{\partial p}{\partial x}=-4x-2y+50=0,$$
$$\frac{\partial p}{\partial y}=-2x-4y+48=0.$$
This linear system of equations has only one solution, $x=26/3$, $y=23/3$, leading to $p=600+2/3\sim 601$. To verify that this is indeed a maximum and not a saddle point or a minimum, we can take second partial derivatives: $\frac{\partial^2p}{\partial x^2}=-4$ and $\frac{\partial^2p}{\partial y^2}=-4$ are both negative. Finally, for these solutions, $x+y=49/3<24$, therefore the inequality is satisfied.
A: $P(x,y)=200+x(50-x)+y(48-y)-(x+y)^2$ 
is the function  that you want to extremize subject to $x+y=24$. I would solve for $x$ or $y$ in the constraint equation, plug it into $P$, differentiate, set equal to zero and solve.
