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Given the function $f\left(x,y\right)=9x^4+16x^3+6x^2y^2$,

defined for all points $(x,y)$ in the area R given by the inequality

$x^2+2x+y^2 \le 0.$

a) Find and classify critical points for $f$ inside of R.

Answer: $\left(-\frac{4}{3},\:0\right)$ Local minimum

I have managed to find this one using the gradient of the function, and setting each part of the gradient equal to zero. Then solve for X,Y to get the points and use the second derivative test to figure if it is a minimum or maximum.

What I don't understand is, when I plug the point to see where it lies on the graph it seems to just be floating in the middle of the air:

Floating point in the middle of the air

b) Decide the absolute maximum and minimum for $f$ on the area R.

Answer: $16, \frac{-256}{27}$

I am not sure what the question exactly asks me to do. Please keep in mind that sentence is translated from another Language to the English language so it might not be 100% correct in this context so feel free to use your imagination of what it could mean.

I realized I could find the absolute maximum by using the technique langrange multiplier with $x^2+2x+y^2 = 0$.

I tried doing the same to find the absolute minimum but setting the function to $x^2+2x+y^2 = -1$, but this seems to be wrong.

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  • $\begingroup$ For $a)$, note the constraint is the circle: $(x+1)^2+y^2\le 1$, the point $(-4/3,0,-256/3)$ floats in air because the graph of a two variable function is a surface. $\endgroup$ – farruhota Aug 7 '17 at 15:56
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Let $x=-\frac{4}{3}$ and $y=0$.

Hence, the condition is valid and $f\left(-\frac{4}{3},0\right)=-\frac{256}{27}$.

We'll prove that $-\frac{256}{27}$ it's a minimal value of our function.

Indeed, we need to prove that $$9x^4+16x^3+6x^2y^2\geq-\frac{256}{27},$$ for which it's enough to prove that $$9x^4+16x^3+\frac{256}{27}\geq0,$$ which follows from AM-GM: $$9x^4+16x^3+\frac{256}{27}=3\cdot3x^4+\frac{256}{27}+16x^3\geq$$ $$\geq4\sqrt[4]{(3x^4)^3\cdot\frac{256}{27}}+16x^3=16|x^3|+16x^3\geq0.$$

Now, $f(-2,0)=16$.

We'll prove that it's a maximal value of $f$.

Indeed, we need to prove that $$9x^4+16x^3+6x^2y^2\leq16$$ and since $y^2\leq-x^2-2x$, it's enough to prove that $$9x^4+16x^3+6x^2(-x^2-2x)\leq16$$ or $$(x+2)(3x^3-2x^2+4x-8)\leq0.$$

But the condition gives $x^2+2x\leq-y^2\leq0.$

Thus, $-2\leq x\leq0$ and $$(x+2)(3x^3-2x^2+4x-8)=(x+2)(3x^3-2(x^2-2x+4))\leq0$$ and we are done!

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