if $x^2+y^2=1$, find the maximum value of $2x^3+y^3$

Question

I tried some things like:

\begin{cases} x^2+y^2=1 \\ 2x^3+y^3=t \end{cases}

Finding the derivative: $t'=6x^2+3y^2=3x^2+3(x^2+y^2)=3x^2+1>= 0$ for every $x$. Therefore, $t$ is always growing.

Then I tried \begin{cases} y=\sqrt{1-x^2}\\ 2x^3+(\sqrt{1-x^2})^3=t \end{cases} but that doesn't make the problem more simple.

I also considered $ t=x^3+x^3+y^3=x^3+(x+y)(x^2+y^2-xy)=x^3+(x+y)(1-xy)$ but that way of solving doesn't make the problem easier either. The answer is: $t_{max}=2$ when $x=1$ and $y=0$. Any ideas?

Answer 1

It's sufficient to consider $x,y$ positive for the maximum. Since $x^2+y^2=1$, we have $x^3\leq x^2\leq x\leq 1$ and $y^3\leq y^2\leq y\leq 1$. Hence $$ 2x^3+y^3\leq x^3 + x^3 + y^3 \leq x^3 + (x^2+y^2) \leq 1+1 = 2 $$ So $$ 2x^3+y^3\leq 2 $$ and taking $x=1$ and $y=0$ you find the equality and the maximum.

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The sobstitution $x=\cos t$, $y=\sin t$ does not work very well as I suggest in comment.

The problem is equivalent to maximize the function $$ f(t)= 2\cos^3 t + \sin^3 t $$ It's derivative is $$ f'(t)=3\sin t \cos t (\sin t - 2\cos t) $$ But the computation for $\sin t-2\cos t =0$ is not simple in my opinion.

Edit: As Amd noticed in comment, the computation $\sin t-2\cos t =0$ just involve more works but is not difficult as I said. And we obtain the results $\sin t = \pm \frac{2}{\sqrt 5}$ and $\cos t = \pm \frac{1}{\sqrt 5}$.

Answer 2

It is a constrained problem and is to be solved using KKT or similar method: the problem is: $max 2*x^3+y^3$ subject to $x^2+y^2=1$ then solve using KKT (or lagrange multiplier)

Answer 3

As hinted in the comments: since $\;x^2+y^2=1\;$ , you can put $\;x=\cos t\,,\,\,y=\sin t\;$ , for some $\;t\in[0,2\pi)\;$, and then you have to find the maximum of

$$f(t)=2\cos^3t+\sin^3t\implies f'(t)=-6\sin t\cos^2t+3\sin^2 t\cos t=3\sin t\cos t\left(\sin t-2\cos t\right)$$

and from here you get

$$f'(t)=0\iff \begin{cases}3\sin t\cos t=0\iff\frac32\sin2t=0\iff2t=k\pi\\{}\\\text{or}\\{}\\\sin t=2\cos t\iff\tan t=2\,\text{ (why can we divide by cosine?)}\iff t=\arctan2+k\pi\end{cases} ,\;\;k\in\Bbb Z$$

Well, you take it from here...

Answer 4

The error in your attempt to differentiate $t$ is that you didn't decide what your variable is! You differentiated one term w.r.t. $x$ and the other w.r.t. $y$, when you should pick one and stick to that choice. For example, if we decide to use $x$ as the variable and "imagine" that $y$ really is a function $y=y(x)$, then by the chain rule we get $$ \frac{dt}{dx}=6x^2+3y^2\frac{dy}{dx}. $$

There are many ways to handling this. Let's first do

Implicit differentiation

Let's use the constraint to find a relation between $dy$ and $dx$. Differentiating both sides of $x^2+y^2=1$ the same way gives $$ 2x+2y\frac{dy}{dx}=0, $$ allowing us to solve $dy/dx=-x/y.$ Plugging this into the earlier derivative gives $$ \frac{dt}{dx}=6x^2-3xy=3x(2x-y). $$ This vanishes when either $x=0$ or $y=2x$. If $y=2x$ then $1=x^2+y^2=5x^2$. This gives you the zeros of $dt/dx$. Remember that the circle equation constrains your $x$ to the interval $x\in[-1,1]$, so you need to check the end points as well.

Using a parameter

The unit circle is parametrized by the familiar trig functions, so $x=\cos\alpha$, $y=\sin\alpha$. Writing everything in terms of $\alpha$ gives $$ t=2x^3+y^3=2\cos^3\alpha+\sin^3\alpha. $$ You can differentiate that w.r.t. $\alpha$ and find the extrema in the usual way. You can assume that $\alpha\in[0,2\pi]$. Remember to go back to $(x,y)$ when presenting your answer.

Answer 5

$$2x^3+y^3\leq2x^2+2y^2=2.$$ The equality occurs for $x=1$ and $y=0$, which says that we got a maximal value.