Question

A cable runs along the wall from C to P at a cost of ​$4 per​ meter, and straight from P to M at a cost of ​$5 per meter. If M is 9 meters from the nearest point A on the wall where P​ lies, and A is 33 meters from​ C, find the distance from C to P such that the cost of installing the cable is minimized and find this cost.

Answer 1

Answer:

The minimum cost of installing the cable is $156.

Step-by-step explanation:

We have an optimization problem.

We have to minimize the cost of the cable.

We will use the variable x to express the the length of cable CP and PM, accordingly to the attache picture.

The length of the cable that goes from C to P (let's call it CP) is x.

$\bar{CP}=x$

Then, the length of the cable that goes from P to M (PM) can be calcualted usign the Pithagorean theorem:

$\bar{PM}=\sqrt{(33-x)^2+9^2}$

The cost function Y is:

$Y=4*\bar{CP}+5*\bar{PM}=4x+5\sqrt{(33-x)^2+9^2}$

To optimize this cost funtion we have to derive and equal to 0:

$\dfrac{dY}{dx}=0\\\\\\\dfrac{dY}{dx}=4+5(\dfrac{1}{2})((33-x)^2+9^2)^{-1/2} *(-2)(33-x)\\\\\\\dfrac{dY}{dx}=4+5\dfrac{x-33}{\sqrt{(33-x)^2+81}}=0\\\\\\\dfrac{x-33}{\sqrt{(33-x)^2+81}}=-\dfrac{4}{5}\\\\\\(x-33)=-\dfrac{4}{5}\sqrt{(33-x)^2+81}\\\\\\(x-33)^2=(-\dfrac{4}{5})^2[(x-33)^2+81]\\\\\\(x-33)^2=\dfrac{16}{25}(x-33)^2+\dfrac{1296}{25}\\\\\\\dfrac{25-16}{25} (x-33)^2=\dfrac{1296}{25}\\\\\\9(x-33)^2=1296\\\\\\x-33=\sqrt{\dfrac{1296}{9}}=\sqrt{144}=\pm12\\\\\\x=33\pm12\\\\\\x_1=33-12=21\\\\x_2=33+12=45$

The valid solution is x=21, as x can not phisically larger than 33.

The cost then becomes:

$Y=4*\bar{CP}+5*\bar{PM}=4x+5\sqrt{(33-x)^2+9^2}\\\\\\Y=4*21+5\sqrt{(33-21)^2+81}\\\\Y=81+5\sqrt{144+81}\\\\Y=81+5\sqrt{225}\\\\Y=81+5*15\\\\Y=81+75\\\\Y=156$