The demand function for a product is given by $p(n)=580-0.8 n^{2}$, where $p$ is the price in dollars and $n$ is the quantity in thousands. If adverse weather conditions are driving the price up by $\$ 5$ per week, find the rate at which the demand quantity for the product is changing when the price is $\$ 260$. The demand for this product is Select an answer $\vee$ by thousand items per week. Select an answer Ouestion Help: [D] Video falling y thousand items per

Step 1 ：The demand function for a product is given by \(p(n)=580-0.8 n^{2}\), where \(p\) is the price in dollars and \(n\) is the quantity in thousands.

Step 2 ：Adverse weather conditions are driving the price up by \$5 per week.

Step 3 ：We are asked to find the rate at which the demand quantity for the product is changing when the price is \$260.

Step 4 ：This is a problem of related rates. We know that the price is increasing at a rate of \$5 per week, and we want to find the rate of change of the quantity.

Step 5 ：We start by differentiating the demand function with respect to time. This will give us an equation that relates the rate of change of the price and the rate of change of the quantity.

Step 6 ：The derivative of the demand function with respect to \(n\) is \(dp/dn = -1.6n\).

Step 7 ：We can then substitute the given values into this equation to find the rate of change of the quantity.

Step 8 ：Let \(dp/dt = 5\) and solve for \(dn/dt\) in the equation \(1.6*dn/dt*n + 5 = 0\).

Step 9 ：Solving the equation gives \(dn/dt = 0.116059586360657\).

Step 10 ：Final Answer: The rate at which the demand quantity for the product is changing when the price is \$260 is approximately \(\boxed{0.116}\) thousand items per week.