Question

Suppose the Cookie Factory has a profit function given by P(n) = 0.02n? + 6n - 21, where P(n) is thetotal profit, in thousands of dollars, from selling and producing n thousands of cookies.How many cookies should be sold to maximize profits?Round to the nearest whole cookie.What is the maximum profit that can be expected?Round to the nearest cent.

Answer 1

The Solution.

The given profit function is

`P(n)=-0.02n^2+6n-21`

To maximize the factory's profit, we have to differentiate the given function with respect to n, and then equate to zero.

`(dP)/(dn)=-2(0.02)n^(2-1)+6n^(1-1)+0=0`
`(dP)/(dn)=-0.04n^{}+6=0`
`-0.04n+6=0`

Solving for x, we get

`-0.04n=-6`

Dividing both sides by -0.04, we get

`n=(-6)/(-0.04)=150\text{ (thousands of cookies)}`

So, the number of cookies to be sold to maximize profit is 150 thousands of cookies.

To get the maximum profit that can be expected, we shall substitute 150 for n in the given profit function. That is,

`P(150)=-0.02(150)^2+6(150)-21`
`P(150)=-0.02(22500)+900-21`
`P(150)=-450+900-21`
`P(150)=429\text{ thousands of dollars}`

Thus, the maximum profit to be expected is 429 thousands of dollars.