I5 Points] DETAILS MY NOTES Suppose that the weekly sales volume \( y \) (in thousands of units sold) depends on the price per unit (in dollars) of the product according to the following formula \( y=32(3 p+1)^{-2 / 5}, p>0 \) (a) What is the rate of change in sales volume when the price is \( \$ 21 \) ? (Round your answer to three decimal places.) \( \frac{d y}{d p} \) (b) Interpret your answer to part (a). (Round your answer to the nearest whole number.) If the price increases \( \$ 1 \), the sales volume will decrease by

**(a)** We are given \[ y = 32(3p+1)^{-2/5}. \] Let \( u = 3p+1 \) so that \( y = 32u^{-2/5} \). Using the chain rule, the derivative is \[ \frac{dy}{dp} = 32\left(-\frac{2}{5}\right)u^{-2/5-1}\frac{du}{dp}. \] Since \[ \frac{du}{dp} = 3, \] we have \[ \frac{dy}{dp} = 32\left(-\frac{2}{5}\right)(3p+1)^{-7/5}\cdot 3. \] Simplify the expression: \[ \frac{dy}{dp} = -\frac{192}{5}(3p+1)^{-7/5}. \] Now, when \( p = 21 \), \[ 3p+1 = 3(21) + 1 = 64. \] Thus, \[ \frac{dy}{dp}\bigg|_{p=21} = -\frac{192}{5}(64)^{-7/5}. \] To compute \( (64)^{-7/5} \), note that \[ 64 = 2^6, \] so \[ 64^{-7/5} = (2^6)^{-7/5} = 2^{-42/5}. \] Numerically, we compute: \[ \frac{192}{5} = 38.4. \] Also, using an approximate value, \[ 64^{7/5} \approx 337.92 \quad \Longrightarrow \quad (64)^{-7/5} \approx \frac{1}{337.92} \approx 0.00296. \] Thus, \[ \frac{dy}{dp}\bigg|_{p=21} \approx -38.4 \times 0.00296 \approx -0.114. \] **(b)** The derivative \(\frac{dy}{dp} \approx -0.114\) tells us the instantaneous rate of change in sales volume \(y\) (in thousands of units) with respect to the price \(p\) at \(p = 21\). That is, for a small increase in price: \[ \Delta y \approx -0.114 \, (\text{in thousands}). \] Since \(y\) is measured in thousands of units, a \( \$1 \) increase in the price will result in a decrease of approximately \[ 0.114 \times 1000 = 114 \] units in weekly sales volume.