Lagrange Question

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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 $468.75$, and then I need to add the $200$ points. So my answer would be $668.75$ which would round to $669$. But this is not correct. What am I doing wrong?

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4
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A few things:

First of all (and most importantly), you call this a Lagrange problem (and your solution does satisfy x + y = 24), but is it really? Seems improbable that the optimal strategy would involve no sleep. I'd look in the interior, $x + y < 24$.

Second, I don't get the same numbers you get. I get the same x and y values (using Lagrange, though these are not optimal), but not the same test score. Are we using the same score function? Mine is: $$ 200+ x(50-x)+y(48-y)-(x+y)^2$$ Using the values you get from Lagrange, I see that as 512.5 . Am I missing something?

4
On

the function you are inserted in is $f(x,y) = 200 +x(50-x) +y(48-y) - (x+y)^2$take the derivative for $x$ and $y$ and set is zero. This will give you the condition for a local extremum. The answer should be $x = 26/3,\ y = 23/3$. Then you still have to explain why that this is a local maximum and not a minimum. And at last you should check the global points.

The maximum test score is: $1802/3 \approx 600$

0
On

fx = 50-4x-2y and fy = 48-4y-2x

50-4x-2y=48-4y-2x

y=x-1

x+y=24 hours, x+(x-1)=24 SO, x= 25/2

y= 24-25/2 = 23/2 ... f(25/2,23/2)= 200+(25/2)*(50−(25/2))+(23/2)(48−(23/2))−((25/2)+(23/2))^2 =512.5

IS IT CORRECT?