Find the supply function $x = f ( p )$ that satisfies $\frac { d x } { d p } = p \sqrt { p ^ { 2 } - 25 }$ and the initial condition x = 700 when $p = \$ 13$ .
A) $x = \frac { 1 } { 3 } \left( p ^ { 2 } - 25 \right) ^ { 3 / 2 } + 124$
B) $x = \frac { 1 } { 3 } \left( p ^ { 2 } - 25 \right) ^ { 1 / 2 } + 696$
C) $x = \frac { 1 } { 3 } ( p - 5 ) + 124$
D) $x = \frac { 1 } { 5 } \left( p ^ { 2 } - 25 \right) ^ { 3 / 2 } + 127$
E) $x = \frac { 1 } { p - 1 } \left( p ^ { 2 } - 25 \right) ^ { 1 / 2 } + 699$

Question

Quizplus · Accepted Answer

To find the supply function $x = f(p)$ given $\frac{dx}{dp} = p\sqrt{p^2 - 25}$, we integrate the given derivative with respect to $p$. The integral of $p\sqrt{p^2 - 25}$ with respect to $p$ is $\frac{1}{3}(p^2 - 25)^{3/2} + C$, where $C$ is the constant of integration. To find $C$, we use the initial condition $x = 700$ when $p = 13$. Substituting these values into the integrated function gives $700 = \frac{1}{3}(13^2 - 25)^{3/2} + C$. Solving for $C$ gives $C = 124$. Therefore, the supply function is $x = \frac{1}{3}(p^2 - 25)^{3/2} + 124$.

Find the Supply Function x=f(p)x = f ( p )x=f(p) That Satisfies

Find the Supply Function $x = f ( p )$ That Satisfies