A cylinder begins with a diameter of 28 yards and a height of 22 yards. If the diameter increases at an instantaneous rate of $2\mathrm{yd}/\sec$ and the height decreases at an instantaneous rate of $-5\mathrm{yd}/\sec$, determine the rate at which the surface area of the cylinder is changing.

Step 1 ：We are given a cylinder with a diameter of 28 yards and a height of 22 yards. The diameter is increasing at a rate of 2 yards per second and the height is decreasing at a rate of 5 yards per second. We are asked to find the rate at which the surface area of the cylinder is changing.

Step 2 ：The surface area of a cylinder is given by the formula $A=2\pi r(r+h)$, where $r$ is the radius and $h$ is the height.

Step 3 ：We can differentiate the surface area formula with respect to time to get an expression for the rate of change of the surface area. This will give us a formula in terms of the rates of change of the radius and the height.

Step 4 ：Taking the derivative, we get $\frac{dA}{dt}=2\pi \frac{dh}{dt}r+\frac{dr}{dt}(2\pi r+2\pi (h+r))$.

Step 5 ：We substitute the given values into the formula: $r=14$, $h=22$, $\frac{dr}{dt}=2$, and $\frac{dh}{dt}=-5$.

Step 6 ：Substituting these values in, we find that $\frac{dA}{dt}=60\pi$.

Step 7 ：So, the rate at which the surface area of the cylinder is changing is $\boxed{{\displaystyle 60\pi {\text{yd}}^2/\text{sec}}}$.