Question

You have noticed that paying higher wages attracts more productive employees. However, you are concerned that there may be a limit to this relationship. Some experimentation has convinced you that the relationship between daily wages paid (x) and profits can be modelled by the equation Profit = 50x − 0.5x2 + .001x3 + 200. The range of wages you are willing to consider is from $0 to $500 per day. Determine the level of wages that will maximize profit.

Answer 1

The level of wages that maximize tge profit is $61.257

And the correspond value for the profit is:

`p(61.257)=50(61.257)-0.5(61.257)^2 + .001(61.257)^3 + 200=1616.502`

Step-by-step explanation:

For this case we have the following function:

`p(x)= 50x -0.5x^2 +0.001x^3 +200`

Where x represent the daily wages paid
`0 \leq x \leq 500`, and p(x) the profit, the objective is maximize this function, and in order to do this the first step is derivate the function respect to x and we got this:

`(dp)/(dx)=50-x+0.003x^2`

As we can see we have a quadratic equation now we need to set up equal the derivate obtained to 0 and then solve for the critical points, like this:

`(dp)/(dx)=0.003x^2 -x +50 =0`

We can use the quadratic formula given by:

`x =(-b \pm √(b^2 -4ac))/(2a)`

And for this case a=0.003 , b=-1 , c =50

Replacing this we got :

`x =(-(-1) \pm √((-1)^2 -4(0.003)(50)))/(2(0.003))`

`x = (1 \pm (√(10))/(5))/(0.006)`

And we got:

`x_1 =61.257 , x_2= 272.076`

Now we need to find the second derivate, like this:

`(d^2p)/(dx^2)=0.006x-1`

And we can replace the values obtained:

`0.006(61.257)-1 =-0.632 <0`

So then 61.257 is a maximum.

`0.006(272.076)-1 =0.632 >0`

So then 272.076 is a minimum.

So then the level of wages that maximize tge profit is $61.257

And the correspond value for the profit is:

`p(61.257)=50(61.257)-0.5(61.257)^2 + .001(61.257)^3 + 200=1616.502`