a) (0,1400) in 2010
dW/dt = (1/25)[W-300]
use differentiation to get ∫ dW/(W-300) = ∫ (1/25)dt
ln|W-300| = (1/25)t + C plug in (0,1400) to find C
ln (1400-300) = C --> C = ln (1100)
ln |W-300| = (1/25)t + ln (1100)
|W-300| = 1100e^((1/25)t)
W = 1100e^((1/25)t) + 300
plug in t= 1/4
W = 1100e^[(1/25)•(1/4)] + 300
W = 1100e^(1/100) + 300
W = 1411
b) use the second derivative which is (1/25)dW/dt to get
(1/25) [1/25(W-300)]
solve for W and you get 300
do a number line and you'll see that anything below 300 is negative (concave down) and anything above 300 is positive (concave up)
so the second derivative of 1/4 would be negative therefore it is an overapproximation