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I am working on a maximum and minimum problem, see picture below. The book video says that the max value of $$f(1,y)=y^{2}+y+5$$ is $(1,1).$ How can I solve this function and get $(1,1)$, as its root is imaginary. this doesn't make any sense to me. Please let me know why. thanks.

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    The max on a closed interval is either a end point of the interval or a critical point. – 2017-02-04
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    @ kingW3 I just realized that, but however, I updated the picture, would you let me know why on L4 and L2, the min is (1,-1/2) instead of (1,-1)? where is that 1/2 coming from? – 2017-02-04
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    It's a critical point, the derivative is equal to 0 and it's smaller then for $y=-1$ – 2017-02-04
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    @ kingW3 sorry, I am a bit confused. $fx= 2x+2xy, fy = 2y+x^{2}$, if I set them equal to 0. I got (0,0) and $(\sqrt{2},-1)$ – 2017-02-04

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We find $f_x = 2 x + 2 x y$, $f_y = x^2+2 y$.

Now we find where $f_x = f_y = 0$, which gives three points, but two are outside our square, so we only have $(x, y) = (0,0)$.

Now, we need to test the function at this critical point and it gives $f(x, y) = f(0,0) = 4$.

We also have to test the endpoints and find that at $(x, y) = (-1, 1), (1, 1)$, we get $f(x, y) = 7$.

Those are the absolute min and max.

There are other local min and max, but they are not the absolute ones and that is what the solution shows. However, we certainly needed to find those to rule out that they are not absolute ones.

Update

We have four points on the square and "everything" in between those four points that can potentially provide an absolute or local min and max. We must test all of them and the critical points to determine where all the extrema are.

So, lets do one of the points. What we do, is fix the point and then evaluate the function on the other variable. Lets take $L_3$.

We start by fixing $x = 1$ and we substitute this in $f(x, y) = f(1, y) = y^2 + y + 5$.

Now, we find the derivative and set equal to zero to find the extrema.

$f'(1, y) = 2 y + 1 = 0 \implies y = -\dfrac{1}{2}$.

We now test $L_3 = (1, 1)$ and this local critical point $\left(1, -\dfrac{1}{2}\right)$. We find that $f(1, 1) = 7$ and $f\left(1, -\dfrac{1}{2}\right) = \dfrac{19}{4}$.

Can you now proceed with the other three points?

Once you do those, then we compare the results from all of the calculations for absolute min and max, recalling that we have the four points, plus the critical point.

Note: The reason we throw out the critical points $(\pm \sqrt{2}, -1)$ is that $\sqrt{2} > 1$, that is, those points are outside our square.

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    then where is the $\frac{1}{2}$ from the min of L2 and L4 coming from? would you please look at my updated picture? – 2017-02-04
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    We test each extreme point and then the values over the range. So, if we let $y = 1$, then we need to find where the min and max over the range of $-1 \le x \le 1$. Then, we test $y = -1$, then $x = 1$ and then $x = -1$ and the ranges for the other variable. Clear? – 2017-02-04
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    sorry, I still don't have figured out how to obtain the min value for $\frac{1}{2}$? especially from the picture, it is $y=\frac{1}{2}$ not x. I am totally lost. sorry. – 2017-02-04
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    @Tmm: Does that update clarify it? – 2017-02-04
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    Yes, it does. it turns out that I was so tired and forgot that the equation y^2+y+5 is the original equation and I can take derivative to get its minimum. Thanks. – 2017-02-04
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    @ Moo by the way, when I set fx and fy =0, I didn't get (0,0), but I got $(\sqrt{2},-1)$ instead. what did I do wrong? – 2017-02-04
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    You actually get three critical points $(x, y) = (0, 0), (\sqrt{2}, -1) , (-\sqrt{2}, -1)$. What is the value of $\sqrt{2}$? This $x-$value is greater than $1$, which is outside of our square, so we toss it. Same for the third point $-\sqrt{2}$, since it is less than $-1$. Clear? – 2017-02-04
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The only critical point interior to the region is $(0,0)$ and the value there is $4$, as noted.

Any other maximum or minimum will occur along the boundary when $\vert x\vert=1$ or $\vert y\vert=1$. So you check the maximum and minimum values of each of those four functions restricted to the intervals $[-1,1]$ for both $x$ values and $y$ values.

For example, $f(1,y)=y^{2}+y+5$ has a maximum value of $7$ at $(1,1)$ and a minimum value of $4.75$ at $x=-\frac{1}{2}$.

Do that for all four sides of the square and find the minimum and maximum on the square.

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    how did you get $x = -\frac{1}{2}$ – 2017-02-04
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    When $x=\pm1$ one gets $f(\pm1,y)=y^2+y+5$ which has a critical point $y=-\frac{1}{2}$. So $\left(\pm1,-\frac{1}{2}\right)$ could be minimum points of the surface restricted to the square $[-1,1]\times[-1,1]$. Turns out they are not since $f(0,0)=4$ is smaller. The maxima will occur at $(\pm1,7)$. – 2017-02-04
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    Remember, $f_y=2y+x^2=2y+1=0$ implies that $y=-\frac{1}{2}$ is a critical point on the left and right boundaries of the square. – 2017-02-04