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hey guys this is a question from homework and I stuck on it.

Find the absolute maximum and absolute minimum of the function $f(x,y) = xy-7y-49x+343$ on or above $y = x^2$ and on or below $y = 53$.

The following is what I tried:

critical points inside the region:
$$f_x = y - 49$$ $$f_y = x - 7$$
so $(7,49)$ is a critical point with function value $0$.
critical points on $y = 53$:
$$f(x,53) = 4x-28$$
min value on the line: $f(-53^(0.5),53) = -57.12043956$
max value on the line: $f(53^(0.5),53) = 1.120439557$
critical points on $y=x^2$:
$$f(x,x^2) = x^3-7x^2-49x+343\\ f'(x,x^2) = 2x^2-14x-49 $$ from $f'(x,x^2) = 0$ we can get $x = 9.562177826$ or $x = -2.562177826$ with function values respectively $109.7266433$ or $405.7733568$

so the absolute maximum should be $405.7733568$ but this is wrong. So what is the correct way to solve it?

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It is the correct way to solve it. You just have a mistake in the last derivative. The coefficient of $x^2$ should be $3$ not $2$

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