V=4e^2+9g^3?

Question

where v is the value of the graduate, e is a number of years prior business experience and g is the graduate school grade point average If v=230 find de/dg when g=2

Answer 1

V is a constant
4e^2 + 9g^3 = V
8ede + 27g^2dg = 0
de/dg = -27g^2/8e

4e^2 = V - 9g^3
V = 230
g = 2
4e^2 = 230 - 72
e^2 = 39.5
e = sqrt(39.5)

de/dg = -27g^2/8e
= -27*2^2/8sqrt(39.5)
= -2.148 (approx)

Answer 2

230 = 4e^2 + 9g^3
Take a derivative with respect to g.
0 = 8e(de/dg) + 27g^2
When g = 2,
230 = 4e^2 + 9(2^3)
230 = 4e^2 + 72
158 = 4e^2
e^2 = 79/2
e = sqrt(79/2)

Plug these in to the formula found after differentiating.
0 = 8sqrt(79/2)*(de/dg) + 27(4)
de/dg = -108/(8sqrt(79/2)) = -27/(sqrt(158)) = -2.148

Answer 3

V=4e^2+9g^3
4e^2 = V- 9g^3
e = .5sqrt(V-9g^3)
de/dg = (.25(V-9g^3)^-.5)*-27g^2
de/dg = (.25(230- 9*8)^-.5) *-27*4
de/dg = -27/sqrt(158) approximately = -2.148 when g = 2.