Let $C(A, B)$ denote a truncated cylinder whose left edge has length $A$ and whose right edge has length $B$. (I'm assuming that it's placed so that its longest and shortest edges are to the left and right).
The volume of $C(A, 0)$ is $\frac{1}{2} \pi r^2 A$. Reason: Stack up $C(A, 0)$ atop $C(0, A)$ to get a cylinder of height $A$. The volume of $C(A, 0)$ is clearly half the volume of this cylinder.
Now for a cylinder C(A, B), where $A > B$, consider the stack
C(0, Q)
C(A, B)
C(B, A)
C(Q, 0)
where $Q = A - B$. THat's a cylinder of height $A + B + (A-B) = 2A$. If we subtract from it the volumes of the top and bottom pieces (which we computed in part 1), we get
$$
\pi r^2 (2A) - \pi r^2 (Q) = \pi r^2 (2A - (A-B)) = \pi r^2 (A + B)
$$
But the volume we've just computed is twice the volume of the sliced cylinder $C(A,B)$ that we're interested in. So the volume of $C(A,B)$ isjust
$$
\pi r^2 \frac{A + B}{2}
$$
A rather similar proof works for surface area.
Note that the proof given above works for a cylinder cut by ANY two planes, as long as the longest and shortest "line" of the truncated cylinder are diametrically opposite each other.
The tank is a truncated circular cone with a base of radius $r=3\,$m, a depth of $4\,$m and top radius of $r=4\,$m. Let $y$ denote the vertical distance from the bottom of the tank. Then $r=3+\frac{1}{4}y$ is the radius of the slice of water lying $y$ meters above the bottom of the tank.
Think of that slice of water as being solid rather than liquid, as if it were frozen. The amount of work done in pumping the water to a height of $5\,$m above the base is the same as if one had to move all those frozen slices to that height.
The volume of each slice is $\pi r^2\,dy$ with $dy$ representing the thickness. The mass of the slice (in the mks system) is found by multiplying the density $\rho=1000\,$ kg/m$^3$ times the volume, so
\begin{equation}
M=1000\pi r^2\,dy=1000\pi\left(3+\frac{1}{4}y\right)^2dy
\end{equation}
Each of these slices of water must be moved upward a distance of $D=5-y$ meters against a gravitational force of $g=9.8\,$m/sec$^2$ resulting in the work for the slice at $y$ being
\begin{equation}
W_y=9800\pi\left(3+\frac{1}{4}y\right)^2(5-y)\,dy
\end{equation}
The total work to remove all the slices of water between $y=0$ and $y=4$ is
\begin{equation}
W=\int_{0}^{4}9800\pi\left(3+\frac{1}{4}y\right)^2(5-y)\,dy
\end{equation}
The rest is routine since the integrand is simply a third degree polynomial. The units will be joules.
Best Answer
See on comparing volume we get to see that volume of cylinder $<$ volume of truncated cone as $r^2<(r^2+R^2-rR)$. Thus from here you can setup equations and differentiate it with radius and then put it to be $0$ for maxima and the condition for maxima is $f''(x)<0$