Clay-modelling

By the Cauchy-Schwarz inequality for integrals, $\left( \int_3^6 \phi(x) \ dx \right) \left( \int_3^6 \frac{\Psi(x)}{\phi(x)} \ dx \right) \ge \left( \int_3^6 \sqrt{\Psi(x)} \ dx \right)^2 = 26^2.$ Then $\int_3^6 \frac{\Psi(x)}{\phi(x)} \ dx \ge 26.$ Equality occurs when $\phi(x) = \sqrt{\Psi(x)}$ .

This problem can be solved with calculus of variations, however, I'm not too familiar with that topic, so I will sketch my solution here:

consider the function $\phi(x)$ at points $a$ and $b$ . The function we want to integrate is $\frac{\Psi(a)}{\phi(a)}$ and $\frac{\Psi(b)}{\phi(b)}$ at those points. If we change the value of $\phi(x)$ by a small $p$ at point $a$ and by $-p$ at point $b$ , the integral of $\phi(x)$ will stay the same, but the integral we want to minimize changes proportionally to $\left(\frac{\Psi(a)}{\phi(a)+p}-\frac{\Psi(a)}{\phi(a)}\right)+\left(\frac{\Psi(b)}{\phi(b)-p}-\frac{\Psi(b)}{\phi(b)}\right)\approx p\left(\frac{\Psi(b)}{(\phi(b))^2}-\frac{\Psi(a)}{(\phi(a))^2}\right)$

Therefore, if $\large \left(\frac{\Psi(b)}{(\phi(b))^2}-\frac{\Psi(a)}{(\phi(a))^2}\right)\neq 0$ we can always make the function smaller by taking away some from the neighborhood of point $b$ and adding it to the neighborhood of point $a$ in $\phi(x)$ , or taking away from $a$ and adding it to $b$ .

Thus, to minimize it, we must have $\large\left(\frac{\Psi(b)}{(\phi(b))^2}-\frac{\Psi(a)}{(\phi(a))^2}\right)=0$ for every $(a,b)$ , which means $\large\frac{\Psi(x)}{(\phi(x))^2}$ is a constant $k$ .

Solving this gives $\phi(x)=\sqrt{\frac{\Psi(x)}{k}}$ .

We have $\frac{1}{\sqrt{k}}\int_3^6\frac{3x}{\sqrt{x-2}}dx=26$

Which gives $k=1$ (integral in the comments), then $\int_3^6\frac{\Psi\left(x\right)}{\sqrt{\Psi\left(x\right)}}dx=\int_3^6\sqrt{\Psi\left(x\right)}dx=26$

Finally, $\lceil{10000\cdot26}\rceil-257400=2600$

I'm sure there must be a simpler way to do this, but calculus of variations works. Start by defining a new function $y(x)=\int_3^x \phi(t) dt$ . The information we have is that $y(3)=0$ and $y(6)=26$ .

The functional we want to minimise is $J[y]=\int_3^6 L(x,y(x),y'(x)) dx$ , where $L(x,y(x),y'(x)) = \frac{9x^2}{y'(x) (x-2)}$ . Note that $L$ does not involve any terms in $y(x)$ , so $\frac{\partial L}{\partial y}=0$ .

Substituting into the Euler-Lagrange equation, we get $\frac{d}{dx} \frac{\partial L}{\partial y'} = 0$ , so that $\frac{\partial L}{\partial y'}$ is a constant.

We have $\frac{\partial L}{\partial y} = -\frac{9x^2}{y'(x)^2 (x-2)} = -\frac{1}{c^2}$ , where the form of the constant has been chosen to make the workings neater.

Rearranging, this becomes $y'(x)=\frac{3cx}{\sqrt{x-2}}$ . Integrating, we find $y(x)=2c(x+4)\sqrt{x-2}+c_1$ , where $c_1$ is another arbitrary constant.

Using the known values of $y(x)$ , we find that $c=1$ ; so $\phi(x)=y'(x)=\frac{3x}{\sqrt{x-2}}=\sqrt{\Psi(x)}$ . It follows that $\mathcal{A}=26$ , and the required answer is $\boxed{2600}$ .