x and y be real numbers

which satisfy $\displaystyle x^2+y^2+xy=1$

then find the minimum value of $\displaystyle x^3y+y^3x$

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- Apr 29th 2010, 07:29 AMbanku12Minimum value
x and y be real numbers

which satisfy $\displaystyle x^2+y^2+xy=1$

then find the minimum value of $\displaystyle x^3y+y^3x$ - Apr 29th 2010, 07:45 AMHallsofIvy
Looks to me like a "Lagrange multiplier" problem!

Let $\displaystyle f(x,y)= x^2+ xy+ y^2$ and $\displaystyle g(x,y)= x^3y+ y^3x$.

For any function f, $\displaystyle \nabla f$= grad f points in the direction of fastest increase so $\displaystyle -\nabla f$ points in the direction of fastest decrease. If there were no constraints, you could find the minimum moving opposite to the way $\displaystyle \nabla f$ was pointing. If you are constrained to surface you cannot necessarily go in that direction but you could still move in the direction of the**projection**of the vector onto the tangent plane of the surface. The only time you could NOT do that is if there is NO such projection which is the same as saying that $\displaystyle \nabla f$ is parallel to the normal vector to the surface.

But for a surface given by g(x,y)= constant, the normal vector is simply $\displaystyle \nabla g$. Saying that one vector is parallel to the other is saying that one is a multiple of the other. The point on g(x,y)= constant that makes f(x,y) a minimum (or a maximum) must satisfy $\displaystyle \nabla f= \lambda\nabla g$ for some constant $\displaystyle \lambda$, the "Lagrange multiplier".

Setting the x and y components of that vector equation equal to each other gives two equations with g(x,y)= constant the third equation you need to solve for the three unknowns, x, y and $\displaystyle \lambda$. Since the value of $\displaystyle \lambda$ is not relevant to the solution, I find that dividing one equation by another, which immediately eliminates $\displaystyle \lambda$ from the equations, is the best way to start. - Apr 29th 2010, 08:33 AMbanku12
could u plz show me, how u found x and y in terms of $\displaystyle \lambda$

i initially tried lik that but cudnt found x and y in terms of $\displaystyle \lambda$