Answer:
(de/dt) = 0.70929
Explanation:
Given
V = 4*e² + 4*g³
where
- V is the value of the graduate
- e is the number of years of prior business experience
- g is the graduate school grade point average
If V = 300 (constant)
g = 2.9
dg/dt = - 0.4
de/dt = ?
we have to get e as follows
V = 4*e² + 4*g³ ⇒ (V/4) = e² + g³ ⇒ e = √((V/4) - g³)
⇒ e = √((300/4) - (2.9)³)
⇒ e = 7.1141
Then we apply
dV/dt = d(4*e² + 4*g³)/dt
d(300)/dt = 4*((2*e*de/dt) + 3*g²*(dg/dt))
0 = (2*e*de/dt) + 3*g²*(dg/dt)
⇒ 0 = 2*(7.1141)*(de/dt) + 3*(2.9)²*(- 0.4)
⇒ (de/dt) = 0.70929