We are going to make p' and n' the respective rates of change of p and n with respect to time.

pn-3p-100n-100=0

Taking derivatives

p'n +n'p -3p'-100n' =0

Given n'= -.250 when n=5 (note n is supply in thousands)

The first equation gives 5p-3p-500-100 = 0

2p =600

p =300

Substituting this value of p and the other givens in the second equation

p'n +n'p -3p'-100n' = 0

5p'-3p' -.250(300)+25 = 0

2p' -75 +25 = 0

2p' = 50

p' = 25 means the price increases by $25 each day (instantaneous)

So p increases as n decreases. To make sense of this if we put in n= 4 instead, we get p = 500. Remember that derivatives are not average rates of change so the relationship is not linear in this case.

Keme L.

Shouldn't n be equal to 5 since in the first line it states " n thousands"? and dN/dT be -0.25 ?02/07/21