# Find absolute maxima and minima of a multivariable function in the domain D

I am trying to find the absolute maxima and minima of $$f(x,y) = 2x^3 + y^4$$ in the domain $$D = \{(x,y)| x^2 + y^2 \le 1 \}$$

I have made an attempt as shown in attached picture. Please ignore dashed areas. Attached picture. The answer is (1,0) and (-1,0). I do NOT want to use the Lagrange multiplier method to solve.

Is it legal to substitute $$y^4 = (1-x^2)^2$$ into $$f(x,y)$$ to make a single variable $$f(x)$$ and then find $$f'(x) =0$$? Then, after finding the $$x$$ can I substitute into $$x^2 +y^2 = 1$$ to find y? Then can I plug it back into $$f(x,y)$$ to find critical points?

I want to know how to find max and min on the boundary $$x^2 + y^2 = 1$$ without the LM method?

• It's not illicit to substitute, I would do the same. – Michael Hoppe Aug 18 at 10:01

Since the origin $$(0,0)$$ is the only critical point inside $$D$$ and the origin is not a local maximum or a local minimum (because $$f(0,0)=0$$ and $$f$$ changes its sign in any neighbourhood of $$(0,0)$$), it follows that the absolute maxima and minima of $$f$$ in the compact set $$D$$ have to be attained at the boundary. Hence, you may consider the restriction of $$f$$ along such boundary $$x^2+y^2=1$$, that is the one-variable function $$g(x)=f(x,\pm\sqrt{1-x^2})=2x^3+(1-x^2)^2$$ in the domain $$[-1,1]$$. Its derivative is $$g'(x)=2x(x+2)(2x-1)$$ so we have to compare $$g(0)=1$$, $$g(1/2)=13/16$$, plus the values at the extreme points $$g(-1)=-2$$ and $$g(1)=1$$. We may conclude that the absolute maximum point is $$(1,0)$$ and the absolute minimum point is $$(-1,0)$$.

Alternative way: if $$x^2 + y^2 \le 1$$ then $$-1\leq x^3\leq x^2$$ and $$0\leq y^4\leq 2y^2$$. Therefore $$f(-1,0)=-2\leq 2x^3\leq f(x,y)=2x^3 + y^4\leq 2x^2+2y^2\leq 2=f(1,0).$$

• @PcumP_Ravenclaw I edited my answer. I hope it can help. – Robert Z Aug 18 at 10:41

Using Calculus, it's routine:

• Find the critical points in the interior of $$D\;$$by setting the partial derivatives of $$f$$ to zero, and solving for $$x,y$$. In this case, the only critical point in the interior of $$D\;$$is the origin, but at the origin, we have $$f(0,0)=0$$, which is not an absolute extreme value of $$f$$ on $$D$$.$$\\[4pt]$$
• On the boundary of $$D$$, you can do what you suggested, namely, replace the $$y^4$$ term of $$f(x,y)$$ by $$(1-x^2)^2$$ and then, using standard methods from single-variable Calculus, proceed to find the absolute extrema of the function $$g(x)=2x^3+(1-x^2)^2$$ on the closed interval $$[-1,1]$$.

But Calculus is not really needed for this problem.

By inspection, we have

• $$f(-1,0)=-2$$.$$\\[4pt]$$
• $$f(1,0)=2$$.

For $$(x,y)\in D$$,

• If $$x \ge 0$$, then $$0 \le 2x^3+y^4 \le 2x^2 + y^2 \le 2(x^2+y^2)\le 2$$.$$\\[4pt]$$
• If $$x < 0$$, then $$-2 \le 2x^3 \le 2x^3+y^4$$.

hence for all $$(x,y)\in D$$, we have $$-2\le f(x,y)\le 2$$.

First suppose $$f(x,y)=-2$$. \begin{align*} \text{Then}\;\;&f(x,y)=-2 \qquad\;\; \\[4pt] \implies\;&x^3+y^4=-2\\[4pt] \implies\;&x^3\le -2\\[4pt] \implies\;&x^3\le -1\\[4pt] \implies\;&x\le -1\\[4pt] \implies\;&x=-1\\[4pt] \implies\;&-2+y^4=-2\\[4pt] \implies\;&y^4=0\\[4pt] \implies\;&y=0\\[4pt] \implies\;&(x,y)=(-1,0)\\[4pt] \end{align*} Next suppose $$f(x,y)=2$$. \begin{align*} \text{Then}\;\;&f(x,y)=2\\[4pt] \implies\;&2x^3+y^4=2\\[4pt] \implies\;&x^3+(x^3+y^4)=2\\[4pt] \implies\;&x^3+(x^2+y^2)\ge 2\\[4pt] \implies\;&x^3+1\ge 2\\[4pt] \implies\;&x^3\ge 1\\[4pt] \implies\;&x\ge -1\\[4pt] \implies\;&x=1\\[4pt] \implies\;&2+y^4=2\\[4pt] \implies\;&y^4=0\\[4pt] \implies\;&y=0\\[4pt] \implies\;&(x,y)=(1,0)\\[4pt] \end{align*} It follows that

• The absolute minimum value of $$f$$ on $$D\;$$is $$-2$$ which occurs for $$(x,y)=(-1,0)$$.$$\\[4pt]$$
• The absolute maximum value of $$f$$ on $$D\;$$is $$2$$ which occurs for $$(x,y)=(1,0)$$.

The maximum and minimum is situated on the curve $$x^2+y^2=1$$ So you have to consider $$f(x,\pm\sqrt{1-x^2})=2x^3+(1-x^2)^2$$ Solve the equation $$6 x^2-4 x \left(1-x^2\right)=0$$ for $$x$$, the first derivative.

• I know that! It could be any point on the unit circle. How do I find them? – PcumP_Ravenclaw Aug 18 at 10:06

$$f(x,y) = 2x^3 + y^4$$ the first derivative: $${df\over dx} = 6x^2 \rightarrow 0; x=0$$ $${df\over dy} = 4y^3 \rightarrow 0; y=0$$ With: $$x^2 + y^2 = 1$$ as an end point. Set each variable to zero and solve for the other. $$x=\pm1$$ $$y=\pm1$$
∴ we have five critical points: $$(0,0) , (0,1), (0,-1), (1,0), (-1,0)$$ plug/substitute each point in the function to get the maximum and minimum.
• Why do you substitute x = 0 and y = 0 seperately into $x^2+y^2 = 1$ ? – PcumP_Ravenclaw Aug 18 at 10:22
$$2x^3+y^4\le 2x^2+y^2$$=$$x^2+x^2+y^2\le x^2+1\le 2$$ . By the same way we get $$2x^3+y^4\ge 2x^3\ge -2$$. We finally check that on the points $$(1, 0)$$ and $$(-1, 0)$$ $$f$$ gets respectively the above mentioned values.