3
$\begingroup$

I need some help with this. I've been struggling through this last chapter of my Calc III class, and I'm not sure how to do this (although, it doesn't seem like it should be difficult to do)

$$ \text{Find the maximum and minimum values of }f(x, y) = 4x + y\text{ on the ellipse } x^{2} + 49y^{2} = 1 \\ \text{Maximum = _____}\\ \text{Minimum = _____} $$

I know that I can find the critical point by taking partial derivatives so $$ \frac{\partial{f}}{\partial{x}} = 4 \\ \frac{\partial{f}}{\partial{y}} = 1 $$ Which gives us the critical point $(4,1)$.

Here's where I get stuck, I know that we can then determine max/min from the equation $D = f_{xx} * f_{yy} - f_{xy}^2$ and based on the value of $D$, we know whether it is a max, a min, or a "saddle" point. It doesn't seem to apply in this situation, because I have an ellipse that I have to use as a constraint.

What do I do next?

UPDATE:

Based on everyone's suggestion, I've been looking into Lagrange multipliers. I think I'm still stuck.

$$ \bigtriangledown{f} = \lambda \bigtriangledown{g} \\ f_x = 4 = \lambda 2 * x\\ f_y = 1 = \lambda 98 * y \\ \text{we need to solve by using ratios of derivatives to remove } \lambda\\ \frac{4}{1} = \frac{\lambda 2 x}{\lambda 98 y} \\ 196 y = x\\ \text{Now we use our ellipse to solve}\\ (196y)^2 + 49y^2 = 1 \\ 38445 y^2 = 1 \\ y = \sqrt{\frac{1}{38465}} = y_0\\ \text{plug y back into our ellipse} \\ x^2 + 49 (\frac{1}{38465}) = 1\\ x = \sqrt{1- \frac{49}{38465}} = x_0 $$ I'm feeling somewhat accomplished, but I have no idea if what I'm doing is correct, or what to do with these new numbers

$\endgroup$
8
  • $\begingroup$ Editing, I was looking at two problems simultaneously and got wires crossed. $\endgroup$ Jul 17 '15 at 10:36
  • $\begingroup$ The point $(0,0)$ does not lie on the ellipse $\endgroup$
    – Henry
    Jul 17 '15 at 10:37
  • $\begingroup$ Do you know Lagrange multipliers ? $\endgroup$ Jul 17 '15 at 10:38
  • $\begingroup$ @ClaudeLeibovici, I've been looking into them based on your suggestion and the hints given in the answers. I've updated my question with my attempt, but I do not think it is correct. $\endgroup$ Jul 17 '15 at 11:51
  • $\begingroup$ @BCqrstoO your answer is correct. Plug the $x_0,y_0$ to $f(x,y)$ to find the extremum. Don't forget that you are solving a quadratic equation... $\endgroup$
    – AnilB
    Jul 17 '15 at 12:03
1
$\begingroup$

You can use lagrange multipliers for this type of question:

$$ L(x, y, \lambda) = (4x + y) -\lambda(x^2 + 49y^2 -1)$$

Solve $\frac{\partial L}{\partial x}=\frac{\partial L}{\partial y} = \frac{\partial L}{\partial \lambda} = 0$

Then substitute the points from the above equations to find maximum and minimum.

If you're not familiar with lagrange multipliers, you can read more here.

here's a quick solution:

$$ 4 -2 \lambda x = 0 $$ $$ 1 - 98 \lambda y = 0$$ $$ x^2 +49y^2 - 1= 0$$

Hence, $$x = 196y$$ $$ (196y)^2 + 49y^2 -1 = 0$$

$$ y = \pm\frac{1}{7\sqrt{785}}$$

$$ x = \pm\frac{28}{\sqrt{785}}$$

$$ (x, y) = \left(\frac{28}{\sqrt{785}}, \frac{1}{7\sqrt{785}}\right), \left(-\frac{28}{\sqrt{785}}, -\frac{1}{7\sqrt{785}}\right)$$

Then substitute these values into $f(x, y) = 4x + y$, clearly, $$ f \left(\frac{28}{\sqrt{785}}, \frac{1}{7\sqrt{785}}\right) = \frac{\sqrt{785}}{7} $$ will be maximum whilst the other point will be minimum.

$\endgroup$
5
  • $\begingroup$ I have given an attempt to use lagrange and updated my question. I do not think I understand what to do $\endgroup$ Jul 17 '15 at 11:51
  • $\begingroup$ Ok, I'll have a look for you. $\endgroup$
    – John_dydx
    Jul 17 '15 at 11:53
  • $\begingroup$ @BCqrstoO, Pls don't forget to upvote or accept answer if its helpful for you. $\endgroup$
    – John_dydx
    Jul 17 '15 at 12:04
  • $\begingroup$ Thanks for going in depth. That last step was what I was missing. I'll be using this as a guide for some of my other problems. Thank you! $\endgroup$ Jul 17 '15 at 12:30
  • $\begingroup$ @BCqrstoO, you're welcome. I just substituted the values of x and y into $4x+y$ in order to find the minimum and maximum values. $\endgroup$
    – John_dydx
    Jul 17 '15 at 13:36
1
$\begingroup$

Lagrange multipliers do work for your problem, but that involves solving three simultaneous non-linear equations in three unknowns. There is another way that uses only one variable: parameterization.

Get a parameterization that describes the given curve in terms of only one variable. In the case of your ellipse you can use

$$x=\cos t, \quad y=\frac 17\sin t, \quad 0\le t\le 2\pi$$

Now you want to want to find the extrema of

$$\hat f(t)=4x+y=4\cos t+\frac 17\sin t$$

There are several ways to find the extrema of that, using calculus or just trigonometry. Here is a calculus way:

$$\hat f'(t)=-4\sin t+\frac 17\cos t=0$$ $$28\sin t=\cos t$$ $$\tan t=\frac 1{28}$$ $$\cos t=\sqrt{\frac 1{\tan^2 t+1}}=\pm\frac{28}{\sqrt{785}}$$ $$\sin t=\sqrt{1-\cos^2 t}=\pm\frac{1}{\sqrt{785}}$$ $$x=\cos t=\pm\frac{28}{\sqrt{785}}$$ $$y=\frac 17\sin t=\pm\frac{1}{7\sqrt{785}}$$

where $x$ and $y$ have the same sign. This means the maximum of $f$ is

$$4x+y=4\frac{28}{\sqrt{785}}+\frac{1}{7\sqrt{785}}=\frac{785}{7\sqrt{785}}=\frac{\sqrt{785}}{7}\approx 4.00255$$

and the minimum is the negative of that. This graph confirms the maximum.

enter image description here

$\endgroup$
0
$\begingroup$

There seems to be a typo in the function $f(x,y)$.

Okay, now that you edited it.

$(4,1)$ is not a critical point. In fact, $f(x,y)$ has no critical points because $f_x$ and $f_y$ is never zero. You should use Lagrange multipliers, not Second partials test.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.