A company that manufactures wigdgets has hired a calculus student to determine the optimal number of widgets to produce each month in order to maximize profit. After a brief review, the student realizes that revenue is determined by

R(n)= 3n^2/e^0.5 +5,
Where n is the number of widgets produced per month, in hundreds, and revenue is measured in millions of dollars.

Determine the number of widgets that should be produced per month to maximize revenue.

is that

R(n)= 3n^2/e^0.5 +5 , as you typed it, or
R(n)= 3n^2/(e^0.5 +5) , or
R(n)= 3n^(2/e^0.5) +5 or .... ?????

I hope I didn't just make it more confusing, its a fraction with the +5 beside it.

To determine the number of widgets that should be produced per month to maximize revenue, we need to find the maximum point of the revenue function. This can be achieved by finding the critical points of the function and then analyzing their behavior.

First, let's find the derivative of the revenue function with respect to the number of widgets produced per month (n).

R'(n) = d/dn [3n^2/e^0.5 + 5]

To do this, we can use the quotient and chain rule:

R'(n) = [(d/dn)(3n^2) * e^0.5 - (3n^2)(d/dn)(e^0.5)] / (e^0.5)^2

R'(n) = [(6n)(e^0.5) - (3n^2)(0.5)(e^0.5)] / e

Simplifying further:

R'(n) = (6ne^0.5 - 1.5n^2e^0.5) / e

Now, to find the critical points, we set R'(n) equal to zero and solve for n:

(6ne^0.5 - 1.5n^2e^0.5) / e = 0

6ne^0.5 - 1.5n^2e^0.5 = 0

Factoring out n:

n(6e^0.5 - 1.5ne^0.5) = 0

This equation can be satisfied if n = 0 or 6e^0.5 - 1.5ne^0.5 = 0.

The first solution, n = 0, does not make sense in this context since we are looking for a positive number of widgets to produce.

For the second solution, 6e^0.5 - 1.5ne^0.5 = 0, we can solve for n:

6e^0.5 - 1.5ne^0.5 = 0
1.5ne^0.5 = 6e^0.5
n = 6/1.5
n = 4

So, we have found one critical point at n = 4.

Now we need to determine whether this critical point is a maximum or minimum. One way to do this is to analyze the second derivative of the revenue function.

Taking the derivative of R'(n) with respect to n:

R''(n) = d/dn [(6ne^0.5 - 1.5n^2e^0.5) / e]

To find the second derivative, we again use the quotient rule:

R''(n) = [(d/dn)(6ne^0.5 - 1.5n^2e^0.5) * e - (6ne^0.5 - 1.5n^2e^0.5)(d/dn)(e)] / e^2

R''(n) = [(6e^0.5 - 3ne^0.5 - 1.5n^2e^0.5) * e - (6ne^0.5 - 1.5n^2e^0.5)(e)] / e^2

Simplifying further:

R''(n) = (6e - 3ne^0.5e - e - 6ne + 1.5n^2e) / e^2

R''(n) = (0.5n^2e - 6ne + 5e) / e^2

Now, we substitute n = 4 into R''(n) to determine its value at the critical point:

R''(4) = (0.5(4^2)e - 6(4)e + 5e) / e^2

R''(4) = (32e - 24e + 5e) / e^2
R''(4) = 13e / e^2
R''(4) = 13 / e

Since e is a positive constant, R''(4) is positive. This indicates that the revenue function is concave up at n = 4.

Therefore, n = 4 corresponds to a local minimum point of the revenue function.

To find the maximum point of the revenue function, we need to consider the behavior of the function as n approaches infinity.

As n goes to infinity, both the squared term (n^2) and the exponential term (e^0.5) become increasingly dominant. The coefficient of the n^2 term is negative (-1.5e^0.5), which means the function will eventually decrease as n approaches infinity.

This means that the maximum point of the revenue function occurs at the critical point n = 4.

Therefore, the optimal number of widgets to produce per month to maximize revenue is 4.

R(n) =

__3n^2___ +5
e^0.5n