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Determine the maximum and minimum value of:

$$x^2+5y^2-4x$$ in the region:

$$x^2+y^2<=1 $$ and $$y>=0$$

I was trying to do this question. Firstly I found the gradient and put it equal to 0 and found the point (2,0). However, this point is not in the region and could not be used. So from here, I'm not quite sure how to continue. I do know it has something to do with the edges of the region, to find the corner points and maybe somehow continue from there, but I'm not quite sure. Thanks

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Your boundary lines are the upper half circle $y=\sqrt{1-x^2}$ with $x\in [-1,1]$ and $y=0$ for $x\in [-1,1]$. So you plug in those values for y into the original equation to turn it into a one variable equation in x, then check for extreme values there by taking the derivative and calculating the function values at every critical point and the endpoints.

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On the boundary $y^2=1-x^2$ for $x\in[-1,1]$ we have the parabola $$x^2+5y^2-4x=x^2+5-5x^2-4x=-4x^2-4x+5$$ with vertex at $x=-1/2$.

Now on the boundaries of $[-1,1]$ there are also extrema.

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$x^2 + 5y^2 - 4x = (x-2)^2 + 5y^2 - 4 \geq -3$ as $x\leq1$ implies $(x-2)^2 \geq 1$ and $5y^2 \geq 0$ and equality happens when $x=1$ and $y=0$. Therefore, minimum value is $-3$.

For maximum value, $x^2 + 5y^2 - 4x \leq x^2 + 5 - 5x^2 - 4x = 6 - (2x+1)^2 \leq 6$ and equality happens when $x = -1/2$ and $y = \sqrt{3/4}$. Therefore, maximum value is $6$.

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