# Maximize the integral of an implicit function with two parameters

**Title.**
Choosing a good title can help to attract interest to your question.
Your original title, "How to use sage to solve this problem?", says
nothing about the question you want to solve. Actually, remembering
we are on ask-sage, it's sort of an empty title. I suggest changing
the title of your question to
"Maximize the integral of an implicit function with two parameters".

**Notation.**
Choosing good notation can help geometric intuition.

The relation you want to study involves $u + y \cos x$ and $v + y \sin x$,
so it would be more natural to use $t$ and $r$ instead of $x$ and $y$,
and to use $x$ to denote $u + r \cos t$ and $y$ to denote $v + r \sin t$.

Reworking the relation to eliminate fractional exponents,
we see that the question is about polar parametrizations
with respect to shifted origins $(u, v)$ of the curve $C$ defined by
$$(1-x^2)^3 = (1-y^3)^2.$$

The questions you are interested in are then:

- for which choices of $(u, v)$ does the relation define $r$ as a function of $t$?
- among those, which one gives the largest average for $r(t)$?

One way you can use Sage to get insight into this question
is by plotting the curve $C$.

```
sage: f(x,y) = (1-x^2)^3 - (1-y^3)^2
sage: implicit_plot(f,(-1.02,1.02),(-0.02,1.3))
```

You see $C$ is a closed curve enclosing a nonconvex domain.

Appropriate choices of an origin $(u, v)$ for a polar parametrization $r(t)$
are those with respect to which the domain enclosed by $C$ is star-shaped.

Given the look of $C$ at $(0,0)$, $(1,1)$, $(-1,1)$, it seems that
$(u, v) = (0,1)$ is in fact the only appropriate choice, and that
no other choice will let you parametrize $C$ by a continuous
function $r(t)$.

Given that, the maximization problem becomes trivial...

Here I made some edits. Hopefully, this could help.

Is there anyway I can solve this problem?

I doubt my question became clearer. It's been months. If u and v has values that make the function discontinuous isn't part of the answer I'm looking for.