Question

attempt 1: 10 attempts remaining. the profit, in dollars, from selling x number of magazines is given by: $p(x) = 2x + sqrt{3x + 11}$ find the rate of change of profit when 200 magazines are sold. (round your answer to the nearest dollar.) the profit is? by? submit answer next item

Explanation:

Step1: Find the derivative of \( P(x) \)

The profit function is \( P(x) = 2x+\sqrt{3x + 11} \). We can rewrite \( \sqrt{3x + 11} \) as \( (3x + 11)^{\frac{1}{2}} \). Using the power rule and the chain rule, the derivative \( P'(x) \) is:
The derivative of \( 2x \) with respect to \( x \) is \( 2 \). For the second term, using the chain rule \( \frac{d}{dx}(u^{\frac{1}{2}})=\frac{1}{2}u^{-\frac{1}{2}}\cdot u' \) where \( u = 3x + 11 \) and \( u'=3 \). So the derivative of \( (3x + 11)^{\frac{1}{2}} \) is \( \frac{1}{2}(3x + 11)^{-\frac{1}{2}}\cdot3=\frac{3}{2\sqrt{3x + 11}} \). Thus, \( P'(x)=2+\frac{3}{2\sqrt{3x + 11}} \)

Step2: Evaluate \( P'(x) \) at \( x = 200 \)

Substitute \( x = 200 \) into \( P'(x) \):
First, calculate the value inside the square root: \( 3x+11=3\times200 + 11=600+11 = 611 \)
Then, \( \sqrt{611}\approx24.72 \)
Now, \( \frac{3}{2\sqrt{611}}\approx\frac{3}{2\times24.72}=\frac{3}{49.44}\approx0.0607 \)
And \( P'(200)=2+\frac{3}{2\sqrt{611}}\approx2 + 0.0607=2.0607\approx2 \) (rounded to the nearest dollar)

Answer:

The rate of change of profit when 200 magazines are sold is approximately \( \$2 \) per magazine.