Find the maximum value of $72\int\limits_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx$

Find the maximum value of $72\int\limits_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx $ for $y\in[0,1].$

I tried to differentiate the given function by using DUIS leibnitz rule but the calculations are messy and I tried to solve directly by integrating it but that also is not working.Can someone please help me in solving this question?

Solution 1:

Let $$ I(y) = \int_0^y \sqrt{x^4 + (y-y^2)^2}dx $$ Let's find $y$ such that $dI/dy=0$. By Leibniz integral rule we have $$ \frac{dI}{dy} = \int_0^y \frac{\partial}{\partial y}\sqrt{x^4 + (y-y^2)^2}dx + \sqrt{y^4 + (y-y^2)^2} $$ But $dI/dy>0$ for $y>0$ ($dI/dy=0$ at $y=0$). So, maximum is reached for $y=1$: $$ I(1)=\int_0^1 x^2\,dx = 1/3 $$

Here is plot of $72I(y)$:

EDIT

(Answer to @PhoemueX's comment)

We should to prove that $dI/dy>0$. We have $$ \frac{dI}{dy} = \int_0^y \frac{(1-2 y) (y-y^2)}{\sqrt{x^4+\left(y-y^2\right)^2}}dx + \sqrt{y^4 + (y-y^2)^2}. $$ It's trivial for $y\le 1/2$ (both terms are positive). For $1/2 < y \le 1$ $$ \frac{dI}{dy} > -\int_0^y \frac{(2 y-1) (y-y^2)}{\sqrt{(y-y^2)^2}}dx + \sqrt{y^4 + (y-y^2)^2} = \\= \sqrt{y^4 + (y-y^2)^2}-(2y-1)y\sqrt{y-y^2}=\\= \sqrt{y^4 + (y-y^2)^2}-\sqrt{y^2(2y-1)^2(y-y^2)} $$ But $$ y^4 + (y-y^2)^2>y^2(2y-1)^2(y-y^2) $$

Solution 2:

As observed by @MichaelGaluza, by Leibniz's rule, it suffices to show $$ \sqrt{y^{4}+\left(y-y^{2}\right)^{2}}+\int_{0}^{y}\frac{\left(1-2y\right)\left(y-y^{2}\right)}{\sqrt{x^{4}+\left(y-y^{2}\right)^{2}}}\,{\rm d}x\geq0. $$ Note that $y-y^{2}=y\left(1-y\right)>0$ for $y\in\left(0,1\right)$. Now, for $y\leq\frac{1}{2}$, we have $\left(1-2y\right)\left(y-y^{2}\right)\geq0$, so that the claim is trivial.

Thus, we can assume $y\in\left(\frac{1}{2},1\right)$. In this case, we have $\left(1-2y\right)\left(y-y^{2}\right)<0$. Furthermore, $$ \int_{0}^{y}\frac{1}{\sqrt{x^{4}+\left(y-y^{2}\right)^{2}}}\,{\rm d}x\leq\int_{0}^{y}\frac{1}{\sqrt{\left(y-y^{2}\right)^{2}}}\,{\rm d}y=\frac{y}{y-y^{2}}=\frac{1}{1-y}. $$ Multiplying by $\left(1-2y\right)\left(y-y^{2}\right)<0$ yields $$ \int_{0}^{y}\frac{\left(1-2y\right)\left(y-y^{2}\right)}{\sqrt{x^{4}+\left(y-y^{2}\right)^{2}}}\,{\rm d}x\geq\frac{\left(1-2y\right)\left(y-y^{2}\right)}{1-y}=\left(1-2y\right)y, $$ so that it suffices to show \begin{align*} & \sqrt{y^{4}+\left(y-y^{2}\right)^{2}}\geq\left(2y-1\right)y\\ \left(\text{since }\left(2y-1\right)y\geq0\text{ because }y>\frac{1}{2}\right)\Longleftrightarrow & y^{4}+\left(y-y^{2}\right)^{2}\geq\left(2y-1\right)^{2}y^{2}\\ \Longleftrightarrow & y^{4}+y^{2}-2y^{3}+y^{4}=y^{4}+\left(y-y^{2}\right)^{2}\geq\left(4y^{2}-4y+1\right)y^{2}=4y^{4}-4y^{3}+y^{2}\\ \Longleftrightarrow & 0\geq2y^{4}-2y^{3}=2y^{3}\left(y-1\right)\\ \left(\text{since }y>0\right)\Longleftrightarrow & y-1\leq0\\ \Longleftrightarrow & y\leq1, \end{align*} which is true. This proves that the derivative is indeed nonnegative, so that the maximum is attained at $y=1$.