Lagrange Multipliers — Question 5

Question 5 – Solution

Strategy. At a constrained extreme with two regular constraints, $\nabla f$ lies in the span of both constraint gradients.

Step 1: Two-multiplier system Let $$g=x^2+y^2+z^2,\qquad h=x+y+z.$$ Then $$\left\langle 0,0,1\right\rangle =\lambda\left\langle 2x,2y,2z\right\rangle+\mu\left\langle 1,1,1\right\rangle.$$ The first two component equations imply $2\lambda(x-y)=0$. If $\lambda=0$, then the first equation gives $\mu=0$, contradicting the third. Hence $x=y=a$.

Step 2: Enforce both constraints The plane gives $z=3-2a$. Substitution into the sphere gives $$2a^2+(3-2a)^2=9,$$ or $$6a(a-2)=0.$$ Thus $a=0$, $z=3$, or $a=2$, $z=-1$. Therefore $$\boxed{\text{highest point }(0,0,3)}, \qquad \boxed{\text{lowest point }(2,2,-1)}.$$

Step 3: Verification Both points satisfy the sphere and plane equations. Their heights are $3$ and $-1$. The intersection is a closed circle, hence compact; the two candidates exhaust the multiplier equations, so these are the absolute height extrema.