a company sells one of its product at Rs.12 per unit a year.its cost function is given by [c(x)=0.01x^2+8^x+7],where x represents the number of units.the number of units and the maximum profit for which profit function maximizes are:

profit = revenue - cost

= 12x - .01x^2 - 8^x - 7

d(profit)/dx = 12 - .02x - (8^x)(ln8)
= 0 for a max of profit

12 - .02x = ln8 (8^x)

if x = 0, LS = 12 , RS = ln8 or appr 2.08
if x = 1, LS = 11.98 , RS = appr 16

so there is a solution between 0 and 1

This question does not make much sense to me
the 8^x term in the cost function does not see right.
as x increases , the 8^x becomes huge.
Why would the cost increase at such a ridiculous rate as your production increases.

check the wording of the question.

To find the number of units and the maximum profit for which the profit function maximizes, we need to calculate the derivative of the profit function and find its critical points.

The profit function can be given by:
P(x) = R(x) - C(x)

Where:
R(x) = Revenue function = Selling price per unit × Number of units sold = 12x
C(x) = Cost function = 0.01x^2 + 8x + 7

Substituting the values in the profit function equation:
P(x) = 12x - (0.01x^2 + 8x + 7)
P(x) = 12x - 0.01x^2 - 8x - 7
P(x) = -0.01x^2 + 4x - 7

To find the maximum profit, we need to find the critical points by taking the derivative of the profit function and setting it equal to zero.

1. Find the derivative of P(x) with respect to x:
P'(x) = -0.02x + 4

2. Set the derivative equal to zero and solve for x:
-0.02x + 4 = 0
-0.02x = -4
x = -4 / (-0.02)
x = 200

The critical point is x = 200 units.

To confirm if this is a maximum or minimum point, we can take the second derivative of the profit function.

3. Find the second derivative of P(x) with respect to x:
P''(x) = -0.02

Since the second derivative is negative (-0.02), this indicates a concave-downward parabola, which implies a maximum point.

Therefore, the number of units for which the profit function maximizes is 200.