a casino hotel manager has approximated demand for two - room suites to be $p = 4000-\frac{q^{2}}{16}$…

a casino hotel manager has approximated demand for two - room suites to be $p = 4000-\frac{q^{2}}{16}$, where $p$ is the price (in dollars) and $q$ is the quantity demanded. use implicit differentiation to find and interpret $\frac{dq}{dp}$ when $q = 4$.\n\nchoose the correct answer below and fill in the answer box to complete your answer.\n\na. $\frac{dq}{dp}=square$, which means that if the price is increased by $1, the quantity demanded decreases by approximately $left|\frac{dq}{dp}\right|$ suites.\n\nb. $\frac{dq}{dp}=square$, which means that if the price is increased by $1, the quantity demanded increases by approximately $left|\frac{dq}{dp}\right|$ suites.

a casino hotel manager has approximated demand for two - room suites to be $p = 4000-\frac{q^{2}}{16}$, where $p$ is the price (in dollars) and $q$ is the quantity demanded. use implicit differentiation to find and interpret $\frac{dq}{dp}$ when $q = 4$.\n\nchoose the correct answer below and fill in the answer box to complete your answer.\n\na. $\frac{dq}{dp}=square$, which means that if the price is increased by $1, the quantity demanded decreases by approximately $left|\frac{dq}{dp}\right|$ suites.\n\nb. $\frac{dq}{dp}=square$, which means that if the price is increased by $1, the quantity demanded increases by approximately $left|\frac{dq}{dp}\right|$ suites.

Answer

Explanation:

Step1: Differentiate both sides with respect to $p$

Given $p = 4000-\frac{q^{2}}{16}$. Differentiating the left - hand side with respect to $p$ gives $\frac{dp}{dp}=1$. Differentiating the right - hand side with respect to $p$ using the chain rule: $\frac{d}{dp}(4000)-\frac{1}{16}\frac{d}{dp}(q^{2})$. Since $\frac{d}{dp}(4000) = 0$, and by the chain rule $\frac{d}{dp}(q^{2})=2q\frac{dq}{dp}$. So, $1 = 0-\frac{1}{16}\times2q\frac{dq}{dp}$.

Step2: Solve for $\frac{dq}{dp}$

From $1=-\frac{q}{8}\frac{dq}{dp}$, we can isolate $\frac{dq}{dp}$ by cross - multiplying. We get $\frac{dq}{dp}=-\frac{8}{q}$.

Step3: Substitute $q = 4$

Substitute $q = 4$ into the formula for $\frac{dq}{dp}$. Then $\frac{dq}{dp}=-\frac{8}{4}=- 2$.

Answer:

A. $\frac{dq}{dp}=-2$, which means that if the price is increased by $$1$, the quantity demanded decreases by approximately $\left|\frac{dq}{dp}\right| = 2$ suites.