Showing max and min of function with constraints

Question

The problem is to find max and min of the function: $$f(x,y)=-16x^6+24x^4-4y^4$$ with constraint $$\sqrt{x^2+y^2} = 1$$ I have been able to transform the problem into a single variable one where $$x=\pm\sqrt{1-y^2}$$ And found the points (-1,0), (1,0). But both of those give the value 8, and I'm not sure how I go about showing if these points are minimum, maximum or saddle points of the function f. Furthermore, can elaborate on how I can make sure the maximum, minimum or saddle point exist.

Answer 1

You can use the method of Lagrange Multipliers.

Let $g(x)=\sqrt{x^2+y^2}$ so that you want to find the min and max of $f$ subject to the constraint $g(x,y)=1$

Next, evaluate $f(x,y)$ at all points $(a,b)$ which satisfy one of the following conditions:

1. $\nabla f(a,b)=\lambda \nabla g(a,b)$ and $g(a,b)=k$

2. $\nabla g(a,b)=(0,0)$ and $g(a,b)=k$
3. $(a,b)$ is an endpoint of the curve $g(x,y)=k$

Here $\nabla f(a,b)$ denotes the gradient vector of $f$ at $(a,b)$.

The max and min value of $f(x,y)$ will be the greatest/least value of $f$ obtained at the points found in steps 1 to 3.

Answer 2

Alternative approach:

Multivariable Calculus is not needed.

You can pretend that the domain is all values $(x,y)$ that are on the unit circle or inside the unit circle, rather than merely all values of $(x,y)$ that are on the unit circle. Then, under this expanded domain, if it turns out that the maximum and minimum values of the original function are achieved at specific values $(x,y)$ that are (specifically) on the unit circle, then the pretense will have been harmless.

So, under the pretended domain, for any fixed value of $x = x_0$, since the only term involving $y$ in the function is $-4y^4$, it is immediate that:

• For $x = x_0$, (i.e., when $x$ is set to the fixed value $x_0$), the function will be maximized at $y^2 = 0$.

• For $x = x_0$, (i.e., when $x$ is set to the fixed value $x_0$), the function will be minimized at $y^2 = 1 - x_0^2$.

Therefore, under the pretended domain, for $|x| \leq 1$ the problem has been reduced to finding the appropriate values of $x$ so that

• $-16x^6 + 24x^4$ is maximized.

• $-16x^6 + 24x^4 - 4\left[(1 - x^2)\right]^2$ is minimized.

From the two bullet points above, without loss of generality, $0 \leq x \leq 1.$

Set $f(x) = -16x^6 + 24x^4 \implies f'(x) = -96x^5 + 96x^3 = 96x^3(-x^2 + 1) \implies $

$\forall x ~$ such that $0 < x < 1$, you have that $f'(x) > 0$, which implies that $f(x)$ is strictly increasing, on this interval.

Therefore, it is immediate that $-16x^5 + 24x^4$ is maximized at $x = 1$. So, the max value of the original function is given by $(x,y) = (1,0)$ which yields a value of $8$.

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For the minimal value of the original function, it is helpful to first analyze:

$$h(t) = -6t^4 + 5t^2 + 1 \implies h'(t) = -24t^3 + 10t = 2t(-12t^2 + 5) ~: ~0 \leq t \leq 1.$$

Here, you have that :

• $h(0) = 1, ~h(1) = 0$.

• $h'(t) = 0 \implies t = 0~$ or $~\displaystyle t = \sqrt{\frac{5}{12}}.$

• $\displaystyle h\left[\sqrt{\frac{5}{12}} ~\right] = \displaystyle -6\left(\frac{5}{12}\right)^2 + 5\left(\frac{5}{12}\right) + 1 > 1.$

So, $h(t)$ must be strictly increasing on the interval $~\displaystyle 0 < t < \sqrt{\frac{5}{12}}~$ and strictly decreasing on the interval $~\displaystyle \sqrt{\frac{5}{12}} < t < 1.$

Therefore, $~\forall t~$ such that $0 < t < 1$, you have that $h(t) > 0.$

Now, to determine the minimum of the original function, analyze the function

$$g(x) = -16x^6 + 24x^4 - 4(1 - x^2)^2$$

$$= -16x^6 + 20x^4 + 8x^2 - 4 \implies $$

$$g'(x) = 16x[-6x^4 + 5x^2 + 1] = 16x \times h(x).$$

Therefore $0 < x < 1 \implies g'(x) > 0.$

Therefore, $g(x)$ is strictly increasing on $0 < x < 1.$

Therefore $g(x)$ is minimized at $x = 0.$

So, the original function is minimized at $(x,y) = (0,1)$ which implies that $-16x^6 + 24x^4 - 4y^2 = -4.$

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Addendum

A shortcut to the 2nd part of the problem just occurred to me.

When attempting to minimize $-16x^6 + 24x^4 - 4\left[(1 - x^2)\right]^2$,

note that by the first part of the problem, you have that the partial sum $-16x^6 + 24x^4$ is strictly increasing on $0 < x < 1.$

Also, it is clear that $- (1 - x^2)^2$ is strictly increasing on $0 < x < 1.$

Therefore, without any additional analysis, based only on the analysis of $f(x) = -16x^6 + 24x^4,$ you know that the overall function must be minimized at $x = 0 \implies y = 1.$