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.

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.

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

The cost function Y is:

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

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

The cost then becomes:

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.

something over 10 so ? /10

step-by-step explanation: