Science:Math Exam Resources/Courses/MATH105/April 2018/Question 03/Solution 1

Let the constraint function $g(x,y)={\frac {1}{3}}y^{2}+2(x-1)^{2}-11$. Note that the objective function is given by $f(x,y)=(x-1)^{2}-y$.

By the method of Lagrange multipliers, if $(x,y)=(a,b)$ is a local extremum of the function $f$, which can be a candidate for the absolute extrema, on the ellipse, then it satisfies

$$f_{x}(a,b)=\lambda g_{x}(a,b),\quad {\text{and}}\quad f_{y}(a,b)=\lambda g_{y}(a,b)$$

for some $\lambda$.

Since we have

$$f_{x}(x,y)=2(x-1),\quad f_{y}(x,y)=-1,\quad g_{x}(x,y)=4(x-1),\quad g_{y}(x,y)={\frac {2}{3}}y,$$

a local extremum $(a,b)$ on the ellipse should satisfy

$${\frac {1}{3}}b^{2}+2(a-1)^{2}=11,\quad 2(a-1)=4\lambda (a-1),\quad -1={\frac {2}{3}}\lambda b.$$

The second equation implies that $(2-4\lambda )(a-1)=0$, so either $\lambda ={\frac {1}{2}}$ or $a=1$should hold.

In the case of $\lambda ={\frac {1}{2}}$, we get $b=-3$ from the third equation. Then, from the first equation, we have $(a-1)^{2}=4$, which is equivalent to $a-1=\pm 2$ and $a=-1,3$. Therefore, possible local extrema for the function $f$ on the ellipse are $(a,b)=(-1,-3)$ and $(a,b)=(3,-3)$.

On the other hand, if $a=1$, then the first equation gives $b^{2}=33$, so that $b=\pm {\sqrt {33}}$. Then, for some $\lambda$ the third equation can be satisfied---we don't need its value. Therefore, in this case, we obtain possible local extrema $(a,b)=(1,{\sqrt {33}})$ and $(a,b)=(1,-{\sqrt {33}})$.

Finally, we compare the function value of $f$ at theses candidates, to find the absolute extrema. Since

$$f(-1,-3)=f(3,-3)=7={\sqrt {49}},\quad f(1,{\sqrt {33}})=-{\sqrt {33}},\quad f(1,-{\sqrt {33}})={\sqrt {33}},$$

on the ellipse, the function $f$ has its absolute maximum value $7$ at $(-1,-3)$ and $(3,-3),$ while it has its absolute minimum value $-{\sqrt {33}}$ at $(1,{\sqrt {33}})$.

Answer: $\color {blue}f(-1,-3)=f(3,-3)=7,f(1,{\sqrt {33}})=-{\sqrt {33}}$