Question

attempt 1: 10 attempts remaining. the total ticket sales ( s(x) ), in dollars, from selling ( x ) tickets is modeled by: ( s(x) = 2x^2 + sqrt{16x + 112} ) find the rate of change of ticket sales when 600 tickets are sold. (round your answer to the nearest dollar.) a) ( s(600) = ) b) the ticket sales is ? by ?

Explanation:

Step 1: Find the derivative of \( S(x) \)

Given \( S(x) = 2x^2+\sqrt{16x + 112} \), we can rewrite the square root term as \( (16x + 112)^{\frac{1}{2}} \).
Using the power rule and the chain rule:

• The derivative of \( 2x^2 \) is \( 4x \).
• For the term \( (16x + 112)^{\frac{1}{2}} \), let \( u = 16x + 112 \), then the derivative of \( u^{\frac{1}{2}} \) with respect to \( u \) is \( \frac{1}{2}u^{-\frac{1}{2}} \), and the derivative of \( u \) with respect to \( x \) is \( 16 \). So by the chain rule, the derivative of \( (16x + 112)^{\frac{1}{2}} \) is \( \frac{1}{2}(16x + 112)^{-\frac{1}{2}}\times16=\frac{8}{\sqrt{16x + 112}} \).

So \( S'(x)=4x+\frac{8}{\sqrt{16x + 112}} \).

Step 2: Evaluate \( S'(600) \)

Substitute \( x = 600 \) into \( S'(x) \):
First, calculate the square root term: \( \sqrt{16\times600 + 112}=\sqrt{9600+112}=\sqrt{9712}\approx98.55 \)
Then, \( 4x = 4\times600 = 2400 \)
And \( \frac{8}{\sqrt{16x + 112}}=\frac{8}{98.55}\approx0.0812 \)
Now, add them together: \( S'(600)=2400 + 0.0812\approx2400.08\approx2400 \) (rounded to the nearest dollar)

Step 3: Interpret the rate of change (for part b)

The rate of change \( S'(600) \approx 2400 \) means that when 600 tickets are sold, the ticket sales are changing at a rate of approximately 2400 dollars per ticket sold. So the ticket sales is increasing by approximately 2400 dollars when the number of tickets sold increases from 600 to 601 (or for a small change in the number of tickets around 600).

Answer:

a) \( S'(600)\approx\boxed{2400} \)
b) The ticket sales is increasing by approximately 2400 dollars.