Hello domenic,
I prefer to work problems like this in general terms, and derive a formula we may then plug our data into. Let's let the base radius of the conical tank be $r$, and the height be $h$.
Let's orient a vertical $y$-axis along the axis of symmetry of the tank, where the origin is at the bottom of the tank, and the initial depth of the fluid in the tank is $y_0$. We wish to find the amount of work $W$ is needed to pump all of the fluid to a point some distance $z$ above the top of the tank.
Now, if we imagine slicing the cone of fluid we wish to remove into disks, we may state, using work is force $F$ times distance $d$ for a constant force:
$$dW=Fd=wd$$
The force exerted is equal to the weight $w$ of the slice.
The weight $w$ of the slice is the product of the weight density $\rho$ in [math]\frac{\text{lb}}{\text{ft}^3}[/math], and the volume [math]V=\pi r_s^2\,dy[/math] of the slice, i.e.:
$w=\rho\pi r_s^2\,dy$
The radius of the slice can be found by similarity.
$$\frac{r_s}{y}=\frac{r}{h}\,\therefore\,r_s=\frac{r}{h}y$$
The distance the slice must be vertically moved against gravity is:
$d=h+z-y$
Putting it all together, we have:
$$dW=\frac{\rho\pi r^2}{h^2}\left(hy^2+zy^2-y^3 \right)\,dy$$
Summing up all the work elements through integration, we obtain:
$$W=\frac{\rho\pi r^2}{h^2}\int_0^{y_0} (h+z)y^2-y^3\,dy$$
Applying the anti-derivative form of the FTOC, there results:
$$W=\frac{\rho\pi r^2}{12h^2}\left[4(h+z)y^3-3y^4 \right]_0^{y_0}=\frac{\rho\pi r^2y_0^3}{12h^2}\left(4(h+z)-3y_0 \right)$$
Now, if $V$ is the volume of the tank, let $kV$ be the initial volume of fluid in the tank, where $0\le k\le1$. To find the initial depth $y_0$ of the fluid in terms of $k$, we may use:
$$\pi\left(\frac{r}{h}y_0 \right)^2y_0=k\pi r^2h$$
$$y_0^3=kh^3$$
$$y_0=h\sqrt[3]{k}$$
Hence:
$$W=\frac{\rho\pi hkr^2}{12}\left(4(h+z)-3h\sqrt[3]{k} \right)$$
We now have a formula into which we may plug the given and known data:
$$\rho=55\,\frac{\text{lb}}{\text{ft}^3}$$
$$h=16\text{ ft}$$
$$k=\frac{1}{2}$$
$$r=8\text{ ft}$$
$$z=3\text{ ft}$$
And so we find for this problem:
$$W=\frac{\left(55\,\dfrac{\text{lb}}{\text{ft}^3} \right)\pi \left(16\text{ ft} \right)\left(\dfrac{1}{2} \right)\left(8\text{ ft} \right)^2}{12}\left(4(16+3)-3(16)\sqrt[3]{\frac{1}{2}} \right)\text{ ft}=\frac{28160\pi}{3}\left(19-6\sqrt[3]{4} \right)\text{ ft}\cdot\text{lb}$$
$$W\approx190959.32\text{ ft}\cdot\text{lb}$$
