# 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

## 3 Answers

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 $$6$$.