A cable runs along the wall from $C$ to $P$ at a cost of $\$ 30$ per meter, and straight from $\mathrm{P}$ to $\mathrm{M}$ at a cost of $\$ 34$ per meter. If $M$ is 16 meters from the nearest point $A$ on the wall where $P$ lies, and $A$ is 100 meters from $C$, find the distance from $C$ to $P$ such that the cost of installing the cable is minimized and find this cost.

Step 1 ：The problem is asking to find the distance from C to P such that the cost of installing the cable is minimized. This is an optimization problem.

Step 2 ：The cost of the cable from C to P is \(30 * CP\) and the cost of the cable from P to M is \(34 * PM\). The total cost is the sum of these two costs.

Step 3 ：We know that \(PM = \sqrt{AP^2 + AM^2}\), where AP is the distance from A to P and AM is the distance from A to M. We also know that \(CP = AP + AC\), where AC is the distance from A to C.

Step 4 ：We can substitute these equations into the cost function and then take the derivative to find the minimum cost.

Step 5 ：Given that AP = AP, AC = 100, AM = 16, CP = AP + 100, PM = sqrt(AP**2 + 256), the cost function is \(cost = 30*AP + 34*sqrt(AP^2 + 256) + 3000\).

Step 6 ：The derivative of the cost function is \(cost_derivative = 34*AP/\sqrt{AP^2 + 256} + 30\).

Step 7 ：Setting the derivative equal to zero and solving for AP, we find that AP_min = -30.

Step 8 ：Substituting AP_min into the cost function, we find that the minimum cost is \$3256.

Step 9 ：Substituting AP_min into the equation for CP, we find that CP_min = 70.

Step 10 ：The minimum cost is \$3256 and the distance from C to P that minimizes this cost is 70 meters. This means that the cable should be installed 70 meters along the wall from C to P, and then straight from P to M. This will result in the lowest cost for installing the cable.

Step 11 ：Final Answer: The distance from C to P that minimizes the cost of installing the cable is \(\boxed{70}\) meters and the minimum cost is \(\boxed{\$3256}\).