# Find max/min value of a multivariable function

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

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.
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.
$$x^2 + 5y^2 - 4x = (x-2)^2 + 5y^2 - 4 \geq -3$$ 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 $$6$$.