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Stationary points 2


Well-known member
Feb 21, 2013
Hello MHB,This is an old exam I got a question,
Decide all stationary points to function \(\displaystyle f(x,y)=x^3y^3\) under constraint \(\displaystyle x^3+y^3+6xy=8\)

So basicly this is how they Solved it and I did so as well but I Also have learned that there is stationary point where gardient of constraint is equal to zero

I am aware that you guys Dont understand but it's nr 3 and are those the stationary points or they forgot when gradient of constraint is equal to zero which gives Also this point \(\displaystyle (0,0)\) and \(\displaystyle (2,-2)\)

Edit:svar means answer so are those point correct or?
\(\displaystyle |\pi\rangle\)


Staff member
Feb 24, 2012
I would use Lagrange multipliers for this problem.

The objective function is:

\(\displaystyle f(x,y)=x^3y^3\)

subject to the constraint:

\(\displaystyle g(x,y)=x^3+y^3+6xy-8=0\)

So, we obtain the system:

\(\displaystyle 3x^2y^3=\lambda\left(3x^2+6y \right)\)

\(\displaystyle 3x^3y^2=\lambda\left(3y^2+6x \right)\)

Solving both for $\lambda$ and equating, we find:

\(\displaystyle \lambda=\frac{x^2y^3}{x^2+2y}=\frac{x^3y^2}{y^2+2x}\)

\(\displaystyle x^2y^3\left(y^2+2x \right)=x^3y^2\left(x^2+2y \right)\)

\(\displaystyle x^2y^2\left(y\left(y^2+2x \right)-x\left(x^2+2y \right) \right)=0\)

\(\displaystyle x^2y^2\left(y^3+2xy-x^3-2xy \right)=0\)

\(\displaystyle x^2y^2\left(y^3-x^3 \right)=0\)

This leads to the same critical points as the solution you have. You don't consider where the gradient of the constraint is zero in such problems.