[CALCULUS RELATED RATES] A supplier produces n thousand boxes of orange daily when it is sold at p dollars per box. Suppose the relationship between n and p is (continuation in description)

Question

A supplier produces n thousand boxes of orange daily when it is sold at p dollars per box. Suppose the relationship between n and p is;

pn-3p-100n-100=0

If the daily supply is decreasing by 250 boxes per day, at what rate is the price changing when the daily supply is 5000 boxes?

Answer 1

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.

Answer 2

We implicitly differentiate this equation: pdn/dt + ndp/dt - 3dp/dt - 100dn/dt = 0;

Because dn/dt = - 250, n = 5000, then 5000p - 3p - 500000 - 100 = 0, 4997p = 500100 and p = 100.08.

100.08(-250) + 5000dp/dt - 3dp/dt - 100(-250) = 0; 4997dp/dt = 20; dp/dt = $0.004 change of price of box per day

Answer 3

p n - 3 p - 100 n - 100 = 0 is the implicit function

n=5

p (5) - 3 p - 100(5) - 100 = 0

5p - 3p - 500 - 100 = 0

2p - 600 = 0

p = 300

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Differentiating....

p dN/dt + dP/dt n - 3 dP/dt - 100 dN/dt = 0

300 * .250 + dP/dt * 5 - 3 dP/dt - 100*.250 = 0

75 + 5 * dP/dt - 3 * dP/dt - 25= 0

50 + 2 * dP/dt = 0

dP/dt = -25