Question

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (ii) f(x,y,z)=y^2-10z subject to the constraint x^2+y^2+z^2=36

Answer 1

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

• If $x=0$, then $L_\lambda=0\implies y=\pm3$.
• If $y=0$, then $L_\lambda=0\implies x=\pm\dfrac32$.
• Either value of $\lambda$ found above requires that either $x=0$ or $y=0$, so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$, $f(0,3)=9$, $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$, and $f\left(\dfrac32,0\right)=\dfrac{729}4$, so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$)

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

• If $x=y=0$, then $L_\lambda=0\implies z=\pm6$.
• If $\lambda=-1$, then $z=-5$, and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$.

We have $f(0,0,-6)=60$, $f(0,0,6)=-60$, $f(0,-\sqrt{11},-5)=61$, and $f(0,\sqrt{11},-5)=61$. So $f$ has a maximum value of 61 and a minimum value of -60.