Can You Find a Function That Satisfies This Integral Equation?

[anemone]

Here is this week's POTW:

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Determine a continuous function $$f:\left[0,\,\frac{1}{3}\right]\rightarrow \left(0,\,\infty\right) $$ with the property such that

$$27\int_{0}^{\frac{1}{3}} f(x) \,dx+16\int_{0}^{\frac{1}{3}} \frac{1}{\sqrt{x+f(x)}} \,dx=3$$-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Here is this week's POTW:

-----

Determine a continuous function $$f:\left[0,\,\frac{1}{3}\right]\rightarrow \left(0,\,\infty\right) $$ with the property such that

$$27\int_{0}^{\frac{1}{3}} f(x) \,dx+16\int_{0}^{\frac{1}{3}} \frac{1}{\sqrt{x+f(x)}} \,dx=3$$-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

Hello all!

I did not realize until I received a PM from Opalg that this week High School Problem Of The Week is a misprint problem and it should read:

Determine a continuous function $$f:\left[0,\,\frac{1}{3}\right]\rightarrow \left(0,\,\infty\right) $$ with the property such that

$$27\int_{0}^{\frac{1}{3}} f(x) \,dx+16\int_{0}^{\frac{1}{3}} \frac{1}{\sqrt{x+f(x)}} \,dx=\frac{21}{2}$$

I am deeply sorry this incident happened again and I sincerely apologize to all the members and staffs of MHB.(Bow)

 

[anemone]

Congratulations to Opalg for his correct solution, which you can find below::)

Spoiler

We are looking for a function $y=f(x)$ such that $$\int_0^{1/3}\Bigl(27y + \frac{16}{\sqrt{x+y}}\Bigr)\,dx = 3.$$ The difficulty seems to be how to make the value of the integral small enough. So I thought that the best way to do this would be to minimise $27y + \dfrac{16}{\sqrt{x+y}}$ for each fixed value of $x$. The minimum of the function occurs when its derivative (with respect to $y$) is zero, namely $27 - \dfrac{16}{2(x+y)^{3/2}} = 0.$ Then $(x+y)^{3/2} = \frac8{27}$, so that $\sqrt{x+y} = \frac23$, and $y = \frac49-x.$ Put this formula for $y$ into the integral, to get the answer $$\int_0^{1/3}\Bigl(27(\tfrac49-x) + \frac{16}{\frac23}\Bigr)\,dx = \int_0^{1/3}(36 - 27x)\,dx = \Bigl[36x - \tfrac{27}2x^2\Bigr]_0^{1/3} = \frac{21}2.$$

The fact that $y$ has to take the least possible value at each point in the interval implies that this function $y = \frac49-x$ is the unique solution to the problem.