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

Answer:

The level of wages that maximize tge profit is $61.257

And the correspond value for the profit is:

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

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:

$\frac{dp}{dx}=50-x+0.003x^2$

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:

$\frac{dp}{dx}=0.003x^2 -x +50 =0$

We can use the quadratic formula given by:

$x =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

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

Replacing this we got :

$x =\frac{-(-1) \pm \sqrt{(-1)^2 -4(0.003)(50)}}{2(0.003)}$

$x = \frac{1 \pm \frac{\sqrt{10}}{5}}{0.006}$

And we got:

$x_1 =61.257 , x_2= 272.076$

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

$\frac{d^2p}{dx^2}=0.006x-1$

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:

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