The demand curve for a product is modeled by \( q=\sqrt{4800-p^{2}} \), where \( p \) is the price per item in dollars, and \( q \) is the quantity in millions. If the company decided to raise the price of the item by \( \$ 3 \) a week, find the rate at which demand changes when the price is \( \$ 40 \). (Round your answer to 2 decimal places.) The demand will be? by

To find the rate at which demand changes with respect to price, we first need to differentiate the demand equation \( q = \sqrt{4800 - p^2} \) with respect to \( p \). The derivative \( \frac{dq}{dp} \) can be computed using the chain rule: \[ \frac{dq}{dp} = \frac{1}{2\sqrt{4800 - p^2}} \cdot (-2p) = \frac{-p}{\sqrt{4800 - p^2}}. \] Next, we evaluate this derivative at \( p = 40 \): \[ \frac{dq}{dp} \bigg|_{p=40} = \frac{-40}{\sqrt{4800 - 40^2}}. \] Calculating the square root: \[ 4800 - 40^2 = 4800 - 1600 = 3200 \quad \Rightarrow \quad \sqrt{3200} = 40\sqrt{8} = 40 \cdot 2\sqrt{2} = 80\sqrt{2}. \] Now substituting back into the derivative: \[ \frac{dq}{dp} \bigg|_{p=40} = \frac{-40}{80\sqrt{2}} = \frac{-1}{2\sqrt{2}}. \] Now we need the rate of change of demand when the price increases by \( \$3 \): \[ \frac{dq}{dt} = \frac{dq}{dp} \cdot \frac{dp}{dt}. \] Given that \( \frac{dp}{dt} = 3 \): \[ \frac{dq}{dt} = \left(\frac{-1}{2\sqrt{2}}\right) \cdot 3 = \frac{-3}{2\sqrt{2}}. \] Calculating that further: \[ \frac{-3}{2\sqrt{2}} \approx \frac{-3}{2 \cdot 1.414} \approx \frac{-3}{2.828} \approx -1.06. \] So, the rate at which the demand changes when the price is \( \$ 40 \) is approximately \( -1.06 \) million units per week. The demand will be decreasing by approximately \( 1.06 \) million units per week.