Decide, using Lagrange multiplier, the extremes for
$\displaystyle f(x,y)=z^3+yz$
where
$\displaystyle y+3z+24=0$
I've tried, but I just couldn't solve the system of equations that arise. Maybe it's z^3 that confuses me.
Decide, using Lagrange multiplier, the extremes for
$\displaystyle f(x,y)=z^3+yz$
where
$\displaystyle y+3z+24=0$
I've tried, but I just couldn't solve the system of equations that arise. Maybe it's z^3 that confuses me.
So far: 2012-09-12 17.02.19.jpg and 2012-09-12 17.02.29.jpg
Attempt:
$\displaystyle 3\cdot \frac{\partial L}{\partial y}-\frac{\partial L}{\partial z}=0\; \Rightarrow\; 3z\left( 1-z \right)=y\; \Rightarrow\; \left( 1,0 \right)\; and\; \left( 0,0 \right).$
The book says $\displaystyle \left( 4,-36 \right)\; and \; \left( -2,-18 \right)$
We are asked to used Lagrange multipliers to find the extrema for:
$\displaystyle f(y,z)=z^3+yz$
subject to the constraint:
$\displaystyle g(y,z)=y+3z+24=0$
By the theorem of Lagrange, we need to solve the system:
$\displaystyle z=\lambda$
$\displaystyle 3z^2+y=3\lambda$
These two equations imply:
$\displaystyle y=3z-3z^2$
Substituting for y into the constraint yields:
$\displaystyle z^2-2z-8=0$
$\displaystyle (z-4)(z+2)=0$
Hence, the extrema occur at:
$\displaystyle (y,z)=(-18,-2),\,(-36,4)$