A cylinder begins with a diameter of 28 yards and a height of 22 yards. If the diameter increases at an instantaneous rate of $2 \mathrm{yd} / \mathrm{sec}$ and the height decreases at an instantaneous rate of $-5 \mathrm{yd} / \mathrm{sec}$, 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 \(\boxed{60\pi \, \text{yd}^2/\text{sec}}\).