THe demand funtion is modeled by D(q) = 56e^-0.000012q, where q is quantity and D(q) is price. What price will yield the maximum revenue? Prove this is a maximum using as qppropriate derivative test.
To determine the price that will yield the maximum revenue, we need to find the derivative of the revenue function and set it equal to zero.
The revenue function is given by the product of quantity and price:
R(q) = D(q) * q
Substituting the given demand function into the revenue function, we have:
R(q) = (56e^(-0.000012q)) * q
Now, we can find the derivative of the revenue function with respect to q. To do this, we can use the product rule:
R'(q) = (D(q) * q)' = D'(q) * q + D(q) * 1
Differentiating the demand function, we get:
D'(q) = (-0.000012 * 56e^(-0.000012q))
Substituting this back into the derivative of the revenue function, we have:
R'(q) = (-0.000012 * 56e^(-0.000012q)) * q + (56e^(-0.000012q)) * 1
Now, we set this expression equal to zero and solve for q:
0 = (-0.000012 * 56e^(-0.000012q)) * q + (56e^(-0.000012q))
Dividing through by (56e^(-0.000012q)), we have:
0 = -0.000012q + 1
Simplifying further, we get:
0.000012q = 1
q = 1 / 0.000012
q ≈ 83,333.33
Now that we have the value of q, we can determine the corresponding price by substituting it into the demand function:
D(q) = 56e^(-0.000012q)
D(83,333.33) ≈ 56e^(-0.000012 * 83,333.33)
D(83,333.33) ≈ 56e^(-0.99999996)
D(83,333.33) ≈ 56 * 0.368
D(83,333.33) ≈ 20.608
Therefore, the price that will yield the maximum revenue is approximately 20.608.
To prove that this is a maximum, we need to use the appropriate derivative test. The second derivative test is often employed to determine if a critical point is a maximum or minimum.
We find the second derivative of the revenue function:
R''(q) = (-0.000012 * 56e^(-0.000012q)) * 1 + (-0.000012 * 56e^(-0.000012q)) * q + (56e^(-0.000012q)) * 0
Simplifying, we have:
R''(q) = 0.000000144q - 0.000012
At the critical point q ≈ 83,333.33, plugging this value into R''(q) gives:
R''(83,333.33) ≈ 0.000000144 * 83,333.33 - 0.000012
R''(83,333.33) ≈ 0.012 - 0.000012
R''(83,333.33) ≈ 0.011988
Since R''(83,333.33) > 0, this indicates that the second derivative is positive, implying that the critical point q ≈ 83,333.33 must be a local minimum.
Therefore, the price that yields the maximum revenue is approximately 20.608, and this is confirmed by the second derivative test.