Find maxima and minima of $f(x,y) = x^3 + y^3 -3x -3y$ in $x + 2y=3$

I need to find max and min of $$f(x,y)=x^3 + y^3 -3x -3y$$ with the following restriction: $$x + 2y = 3$$.

I used the multiplier's Lagrange theorem and found $$(1,1)$$ is the minima of $$f$$. Apparently, the maxima is $$(-13/7, 17/7)$$ but I could not find it via Lagrange's theorem.

Here's what I did:

I put up the linear system:

$$\nabla f(x,y) = \lambda \, \nabla g(x,y)$$

$$g(x,y) = 0$$

then,

$$(3x^2 -3, 3y^2 -3) = \lambda (1,2)$$

$$x + 2y -3 = 0$$

Solving for $$\lambda$$, I got $$\lambda = 0$$, which gave me $$x = 1$$ and $$y = 1$$.

How can I find the maxima if lambda only gives one value which is $$0$$?

• no global maximum or global minimum. One local for each. As a one variable problem, take $x= 1 + 2 t,$ $y = 1 - t$ as a parametrization of the constraint (line), calculate $x^3 + y^3 - 3x - 3y$ in terms of $t$ Dec 2 '20 at 17:45
• "But I need to use Lagrange's theorem...". If I had known that, then I wouldn't have answered. Unfortunately, I'm not familiar with this theorem. I have therefore removed my answer. It would help if (for next time), you try to make your query very clear. Dec 2 '20 at 18:07
• @user2661923 The last line of the question is: "How can I find the maxima if $\lambda$ only gives one value which is $0$?" It is clear that the OP is trying to solve this problem a certain way and got stuck. Dec 2 '20 at 18:26
• @Théophile Yes, that was clear. What was not clear was that he was required to use the method for which he was stuck. Given how off-the-wall many mathSE queries are, re an OP often taking unusual avenues, I was fully justified in answering the query that he originally posed, which is "how do I determine the minimum and maximum values?" Dec 2 '20 at 18:34
• @user2661923 Yes, there are many questions on MSE that are missing context or necessary information. This is not one of them; the OP specifically asked what went wrong with $\lambda$. You chose to take a different avenue because you are unfamiliar with Lagrange's Theorem. Rather than getting upset, why not take this as an opportunity to learn the theorem? :) Dec 2 '20 at 18:43

As others have said, you don't necessarily need to use Lagrange multipliers. But since you've set up the system, we can see what happens:

$$3x^2-3=\lambda\\ 3y^2-3=2\lambda\\ x+2y-3=0$$

From the first two equations, we have $$3y^2-3=2(3x^2-3)$$, which simplifies to $$y^2-1=2(x^2-1)$$. Rearranging the linear constraint, we have $$x=3-2y$$. Putting this information together leads to $$7y^2-24y+17=0.$$

You could solve this using the quadratic formula, but it is quicker to observe that $$-24 = -7-17$$:

$$7y^2-7y-17y+17=0$$

and so $$(7y-17)(y-1)=0$$.

This will give you the two local extrema.

• Right, I found the maxima as (-13/7, 17/7) but if I plug this pair into the hessian function (det = | (6x, 0), (0,6y)| and I get a negative value, which means this point is not local extrema. Why this contradiction? Dec 2 '20 at 18:26
• @Brasilian_student Good question. Keep in mind that the point in question is not actually a critical point of $f(x,y)$ because it does not satisfy $f_x(x,y)=f_y(x,y)=0$. Rather, it is only an extreme point with respect to the restriction $x+2y=3$. On the other hand, $(x,y)=(1,1)$ is an extreme point of $f$. Dec 2 '20 at 18:39

This is from your working -

$$(3x^2 -3, 3y^2 -3) = \lambda (1,2)$$

$$3x^2 - 3 = \lambda, 3y^2-3 = 2\lambda$$

Equating $$\lambda$$ from both equations,

$$6x^2-6 = 3y^2-3 \implies 2x^2 - y^2 = 1$$

Substitute $$x$$ from $$x+2y = 3$$

$$2(3-2y)^2 - y^2 = 1$$

$$\implies 7y^2 - 24y + 17 = 0 \,$$ or $$(7y-17)(y-1) = 0$$

Can you take it from here and find possible points for extrema?

In this case, it is really not a great idea to use Lagrange multipliers. We can write $$x$$ in terms of $$y$$ (or vice-versa) using the restriction and reduce this question to a one variable optimisation problem. Substituting $$x = 3 - 2y$$, the problem reduces to finding the extrema of $$f(y)=-7 y^3+36 y^2-51 y+18$$. The critical points are $$y = \frac{17}{7}$$ and $$y = 1$$ and, considering that $$f''(1)>0$$ and $$f''(\frac{17}{7})<0$$, they are a local minimum and maximum, respectively.

It is also worth noticing that, since $$\displaystyle \lim_{y \to \pm \infty} f(y) = \mp \infty$$, there will not be any global extrema.

• It is useful to leave some comment together with a downvote. I totally accept the down vote, but would be happy to improve the answer. Dec 2 '20 at 17:57
• @AnindyaPrithvi Thanks for the feedback. I'll make some changes to the answer. Regarding the Lagrange multipliers, it was a very conscious option not to use them: I think that this a bad illustration of their use because the results are weaker than changing to a 1d problem. Dec 2 '20 at 22:12