Math Problem Statement

Given the function P=64Q−(200Q−767)1.5600 . Calculate the value of dP/dQ when Q=15 .

Note: Round your answer to 2-decimal places.

Solution

To calculate dPdQ\frac{dP}{dQ} for the function

P=64Q(200Q767)1.56100P = 64Q - (200Q - 767)^{1.56} \cdot 100

we need to differentiate PP with respect to QQ and then evaluate it at Q=15Q = 15.

  1. Differentiate PP:

Using the chain rule and product rule, we find:

dPdQ=641001.56(200Q767)0.56ddQ(200Q767)\frac{dP}{dQ} = 64 - 100 \cdot 1.56 \cdot (200Q - 767)^{0.56} \cdot \frac{d}{dQ}(200Q - 767)

ddQ(200Q767)=200\frac{d}{dQ}(200Q - 767) = 200

Thus,

dPdQ=641001.56(200Q767)0.56200\frac{dP}{dQ} = 64 - 100 \cdot 1.56 \cdot (200Q - 767)^{0.56} \cdot 200

  1. Evaluate at Q=15Q = 15:

First, calculate 200Q767200Q - 767 at Q=15Q = 15:

200(15)767=3000767=2233200(15) - 767 = 3000 - 767 = 2233

Now substitute Q=15Q = 15 into the derivative:

dPdQ=641001.56(2233)0.56200\frac{dP}{dQ} = 64 - 100 \cdot 1.56 \cdot (2233)^{0.56} \cdot 200

Calculating (2233)0.56(2233)^{0.56}:

(2233)0.5616.57(using a calculator)(2233)^{0.56} \approx 16.57 \quad (\text{using a calculator})

Now substitute this value back:

dPdQ=641001.5616.57200\frac{dP}{dQ} = 64 - 100 \cdot 1.56 \cdot 16.57 \cdot 200

Calculating the multiplication:

1001.5616.57200516800100 \cdot 1.56 \cdot 16.57 \cdot 200 \approx 516800

So,

dPdQ=64516800\frac{dP}{dQ} = 64 - 516800

Calculating this:

dPdQ516736\frac{dP}{dQ} \approx -516736

  1. Final Rounded Answer:

Thus, the value of dPdQ\frac{dP}{dQ} when Q=15Q = 15 is approximately:

dPdQ516736.00\frac{dP}{dQ} \approx -516736.00

Summary

The value of dPdQ\frac{dP}{dQ} at Q=15Q = 15 is 516736.00-516736.00.

Would you like to go over any details or have any questions? Here are some related questions for further exploration:

  1. What are the implications of the derivative value in terms of the function's behavior?
  2. How would the derivative change if the exponent were different?
  3. Can you explain the significance of the chain rule in this context?
  4. What happens to the derivative if QQ approaches a different value?
  5. How would you interpret this derivative in a real-world scenario?

Tip: When dealing with derivatives, always ensure to apply the correct differentiation rules to avoid errors in calculation.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differentiation
Calculus
Chain Rule
Product Rule

Formulas

P = 64Q - (200Q - 767)^{1.56}
dP/dQ = 64 - 100 * 1.56 * (200Q - 767)^{0.56} * 200

Theorems

Chain Rule
Product Rule

Suitable Grade Level

Grades 11-12