A cylinder begins with a radius of 11 yards and a height of 30 yards. If the surface area of the cylinder decreases at an instantaneous rate of $-220 \mathrm{yd}^{2} / \mathrm{sec}$ and the height increases at an instantaneous rate of $4 \mathrm{yd} / \mathrm{sec}$, determine the rate at which the radius of the cylinder is changing.
Give your answer as a decimal number. The units are provided after the answer box.
Answer:
$\frac{y d}{\sec }$

Step 1 ：The surface area of a cylinder is given by the formula \(A = 2\pi rh + 2\pi r^2\). We are given that the surface area is decreasing at a rate of -220 square yards per second, and the height is increasing at a rate of 4 yards per second. We want to find the rate at which the radius is changing.

Step 2 ：We can express the rate of change of the surface area with respect to time as \(\frac{dA}{dt} = 2\pi r\frac{dh}{dt} + 2\pi h\frac{dr}{dt} + 4\pi r\frac{dr}{dt}\).

Step 3 ：Substituting the given values, we get \(-220 = 2\pi(11)(4) + 2\pi(30)\frac{dr}{dt} + 4\pi(11)\frac{dr}{dt}\).

Step 4 ：Solving for \(\frac{dr}{dt}\), we get \(\frac{dr}{dt} = \frac{-220 - 2\pi(11)(4)}{2\pi(30) + 4\pi(11)}\).

Step 5 ：Calculating the above expression, we get \(\frac{dr}{dt} = -0.293\) yards per second.

Step 6 ：So, the radius of the cylinder is decreasing at a rate of \(\boxed{0.293}\) yards per second.