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Given a point $(x,y)$ on the curve $2(x^3+y)^4+(x^3+y)^2=2x^3+y$,

Find the maximum value of $y$

Here is my solution of the problem.

enter image description here

I have a doubt as to this solution because two variables, $u$ and $v$ are not independent of each other, but interwound with a relation between $x$ and $y$.
However, it doesn't mean that change of $u$ affects and changes the value of $v$ because there will be an appropriate pair $(x,y)$ for changed $u$ and unchanged $v$. (Also, the domain of $u$ and $v$ is $\mathbb{R}$)

So it's quite confusing whether my solution is valid or not. Is it a valid solution?

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Riddle Aaron
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    The dependency of $(u,v)$ wrt. $(x,y)$ does not matter because the point sets are identical. The transformation is invertible. –  Nov 19 '19 at 09:38
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    @RiddleAaron Your solution is fine... There is even a one-to-one correspondence between the (x,y) and (u,v) variables. – PierreCarre Nov 19 '19 at 09:41
  • Thanks for the comment! – Riddle Aaron Nov 19 '19 at 09:43
  • @PierreCarre Could I ask if it would be fine using typical differentiation rather than Lagrange multiplier method to solve the problem above? – Riddle Aaron Nov 19 '19 at 09:46
  • @RiddleAaron In this case, the restrictions allows you to write $v$ in terms of $u$, so you can reduce this to a one variable optimization. So the answer is yes. – PierreCarre Nov 19 '19 at 10:04

3 Answers3

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Check:

Below a plot of $v=2u^4+u^2-u$ and $v=u-\frac58$. Tangency proves the optimum.

enter image description here

By the way, this graphical representation hints that you have to find a double root of

$$2u^4+u^2-u=u-a,$$

hence the root of the derivative

$$8u^3+2u-1=1,$$ which lets you compute $a$.

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Your solution is fine, as an alternative using implicit differentiation

$$2(x^3+y)^4+(x^3+y)^2=2x^3+y $$

$$\implies 24x^2(x^3+y)^3dx+8(x^3+y)^3dy+6x^2(x^3+y)dx+2(x^3+y)dy=6x^2dx+dy$$

$$(1-2(x^3+y)-8(x^3+y)^3)dy=(24x^2(x^3+y)^3+6x^2(x^3+y)-6x^2)dx$$

$$\frac{dy}{dx}=\frac{24x^2(x^3+y)^3+6x^2(x^3+y)-6x^2}{1-2(x^3+y)-8(x^3+y)^3}=0$$

that is

  • $x=0 \implies 2y^4+y^2-y=0 \implies y=0 \quad \lor \quad y\approx 0.59$

or

  • $4(x^3+y)^3+(x^3+y)-1=0 \implies x^3+y=\frac12 \implies 2\cdot\frac1{16}+\frac14=1-y \implies y=\frac 5 8$
user
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In this case, since the restriction reads $v = 2u^4+u^2-u$, we can just maximize the function $f(u)= u - (2u^4+u^2-u) = -2u^4 - u^2 + 2u$. The maximum is attained at $u=\frac 12$ and $f(\frac 12)=\frac 58$.

PierreCarre
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