Find Minimum

I meant to say this :

consider the graph of the function

x²f(x)-xf²(x)=G(x)

put f(x)=y
now for a particular x this value will will depend only on y

it will be maximum when dG(x)/dy=0 ( partial )

that is when

x²-2xy=0

=> y=x/2

thus at each x the value of this quantity will be maximum when y=x/2

thus if y=x/2 the quantity will be the maximum possible for all x from 0 to 1

thus then the area will also be maximum(as it will be maximum possible at every point)

rohan, why dont you actually write out the steps.

Ok I(f) = \int_0^1 x^2 f(x) - x f^2(x) \ dx

Now \frac{\partial I}{\partial f} = 0

What next?

Rohan : Can u pl. show sum more steps dude i didnt get much......... and iska matlab kya hai??

"if are is to be maximum if"

thus are will aso be maximum in that case

and one error in the last line it will be partial differentiation ..w.r.t f(x)

i am talking for each definite x that varies from 0 to 1 read my soln carefully

u cant hav x fixed it varies .... ye integral hai mere bhai

This is the method. I flicked this one from goiit where i posted this solution about a month back.

Here, the stumbling block is f(x), because it can be just about any function. Then it strikes that you encounter a similar situation when you are looking for the global minimum or maximum of a quadratic

So there you isolate the constant and push the unknown x into a perfect square

Same thing is done here for f(x). Because we know how to integrate a polynomial in x. So we push f(x) into some perfect square and we are rid of it

Problems in JEE will use some crossover techniques like this. When you play around with the expression, your brain will trigger some response that we have encountered this situation before. follow such hunches and you will be through!

prophet sir any alternate soln for this problem plzz...

sumthin including more of calculus than makin perfect sq....

we generlly dont do this in calculus

Let g(x)=x^2f(x)-xf^2(x). Complete the square in f(x) as follows:
g(x)=-x\left(f^2(x)-2\cdot f(x)\cdot \dfrac{x}{2}+\dfrac{x^2}{4}-\dfrac{x^2}{4}\right)
\Rightarrow\ g(x)=\dfrac{x^3}{4}-x\left(f(x)-\dfrac{x}{2}\right)^2\leq \dfrac{x^3}{4}
since 0 ≤ x ≤1.
Therefore,
\dfrac{x^3}{4}-g(x)\geq 0
\Rightarrow\ \int_0^1\left(\dfrac{x^3}{4}-g(x)\right)\mathrm{d}x\geq 0
\Rightarrow\ \int_0^1\dfrac{x^3}{4}\ \mathrm{d}x -\int_0^1g(x)\ \mathrm{d}x\geq 0
\Rightarrow\ \int_0^1g(x)\ \mathrm{d}x\leq \dfrac{1}{16}
Accordingly, the maximum value of \int_0^1 (x^2f(x)-xf^2(x))\ \mathrm{d}x is \dfrac{1}{16}. The equality is attained for f(x)=\dfrac{x}{2}.

did i do anyhting worthful .... or jus blabbered // ?

solving the above diff eqn :

we get ..

f²(x) - xf(x) = x²e^-3x+C

=>x f²(x) - x²f(x) = x³e^-3x+C

so we get

z = x³e^-3x+C

that f² means square or second derivative ?

RHS is a function of what variable?

You are differentiating w.r.t to what variable?

but since the limits are given to be constants

differentiating the entire thing would always give zero