# Finding the Maximum and Minimum Values of a Function in a Domain

Find the maximum and minimum values of the function $$f(x, y) = 2x^2+3y^2-4x-5$$ on the domain $$x^2+y^2\le 225$$.

After finding the first partial derivatives, I found that $$(1, 0)$$ was a critical point and I found that it was a local minimum from the second derivative test. So the minimum value of $$f(x, y)$$ would be $$-7$$ at the point $$(1, 0)$$.

However, what I am confused about is how to find the maximum value and point. Since this function does not have a local maximum point, I thought that the answer would simply be on the boundaries of the inequality. However, it seems that neither the point $$(15, 0)$$ or $$(0, 15)$$ give the correct answer.

If anyone knows how I should approach this problem and can provide feedback, I would be very grateful!

• The boundary of the domain is a full circle defined by $x^2+y^2\leq 225$, not just the two points $(15,0)$ and $(0,15)$. – 高田航 Nov 24 '18 at 6:15
• Yes I see. So in that case, could you please tell me how I should approach this problem instead? Thank you! – sktsasus Nov 24 '18 at 6:31
• have you heard about Langrange multipliers? – Thomas Nov 24 '18 at 6:37

According to the method of Lagrange parameters, the gradient of $$f$$ will be normal to the boundary in an extremal value. The normal to the circle in a given point $$(x,y)$$ is just the position vector $$(x,y)$$, so you have to look for points where $$\nabla f$$ is a mulitple of that vector.

Now

$$\nabla f =\left ( \array{4x-4 \\ 6y} \right)$$ which means we are looking for $$x, y$$ such that $$4x-4 = \lambda x, 6y= \lambda y$$ with the side condition $$x^2+y^2=225$$

This is true iff ($$\lambda = 6$$ or $$y=0$$).

If $$\lambda = 6$$ you easily see that then $$x=-2$$ and $$y$$ is determined by the side condition.

If $$y=0$$ $$x$$ is determined by the side condition (and $$\lambda$$ by the equation for $$x$$ and $$\lambda$$).

Plugging in $$x$$ and $$y$$ into the function will show whether you have a maximum or minimum.

• Thank you for the answer! I am familiar with Lagrange Multipliers but I just wasn't sure how they would work here. But now I do see that I needed to check for the global minimum as well as the critical point I found may not have been it. Thanks for the clarification! – sktsasus Nov 24 '18 at 16:08

As the guy in the comment says, the boundary of the domain is the circumference of the circle $$x^2 + y^2 = 225$$, not just the two points. Since the function has a minima at $$(1,0)$$ and no other critical points in the region, it means that the function is increasing in every direction from that point. So the maxima of the function will be at one of the boundary points. Now if the point is on the boundary, it will lie on the circle's circumference and satisfy the equation $$x^2 + y^2 = 225$$. Putting this in the function equation, we get

$$f(x,y) = 2x^2 + 3y^2 -4x -5$$ $$2x^2 + 3(225 - x^2) -4x -5$$ $$g(x)=-x^2 -4x + 670$$

Now you can diffrentiate to find out where this will have its maxima

$$g'(x) = -2x - 4 = 0$$ $$x=-2$$

As you can quickly confirm from the original equation or the graph of a quadratic equation, this is a maxima. Using this to get $$y$$

$$y^2 = 225 - x^2 = 225 -4 = 221$$ $$y = \sqrt{221}$$

Putting $$x=-2$$ and $$y=\sqrt{221}$$ back in the original function

$$f(x,y) = 674$$

• Note the function must either be constant or must have at least one minimum and one maximum along the boundary, since that is compact and the function is continuous. You are loosing at least one solution by your approach. – Thomas Nov 24 '18 at 6:55
• Actually, $f(x,y) = f(x, -y)$, so with your values of $x, y$ you found two minima or maxima with the same value. This means you actually missed at least two other extrem values. – Thomas Nov 24 '18 at 7:32
• Hint: note that $-25 \le x\le 25$, so also the one dimensional problem to which you reduced the two dimensional one is an extreme value problem with a boundary condition. – Thomas Nov 24 '18 at 7:34
• The values you found for $y$ are the locations of the two maxima. In $y=0, x=\pm 25$ you have two more extreme values (with respect to the boundary), which are the minima of $f$ along the boundary (but not global minima, the global minimum is attained in the interior of the region, which was discoverd by the OP already). – Thomas Nov 24 '18 at 10:04
• @Thomas Every occurrence of "25" should be "15" in your comments. This doesn't affect anything else, but may have misled others. – Teepeemm Nov 24 '18 at 16:07

If you don't know, or want to use, Lagrange Multipliers as in Thomas's answer, and you want to find the local extrema on the boundary to make sure you don't miss anything, then you can either handle the endpoints with $$x=\pm15$$ that Sauhard Sharma neglected to address in their answer, or you can give the circle a more natural parametrization.

Circles can be parametrized with cosines and sines, so the circle $$x^2+y^2=225$$ is parametrized by $$x=15\cos t,\, y=15\sin t$$. If you substitute this in to the objective function $$2x^2+3y^2-4x-5$$, you'll see how the values of the function above the circle depend on the angle $$t$$, and you can then use single-variable calculus to find the local extrema. Since cosine and sine are continuous on the whole real line, you don't have to worry about missing end-points with this method.

With free CAS Maxima and my programm "nopt":

package "nopt"

examples

• @ Mark S Link to package "nopt" I include. – Aleksas Domarkas Nov 26 '18 at 11:29