1) To find the rate of change of the demand, we need to take the derivative of the equation with respect to p.
Differentiating both sides of the equation with respect to p, we get:
4x + 6p + 100p = 0
Solving for x, we get:
x = -3p/2
Substituting p = 30, we get:
x = -45
Therefore, the rate of change of the demand is -45 when the price is 30 dollars and the price is increasing at a rate of 4 dollars per month.
2a) To find A, we need to take the derivative of the equation with respect to q.
Differentiating both sides of the equation with respect to q, we get:
A = 4q
2b) Substituting q = 20 and dp/dt = 4, we get:
dR/dt = 80(4) = 320
1) A price p (in dollars) and demand x for a product are related by
2x^2+6xp+50p^2=10600.
If the price is increasing at a rate of 4 dollars per month when the price is 30 dollars, find the rate of change of the demand.
2)
a) The price (in dollars) p and the quantity demanded q are related by the equation: p^2+2q^2=1100.
If R is revenue, dR/dt can be expressed by the following equation: dR/dt=A dp/dt,
where A is a function of just q.
b)Find dR/dt when q=20 and dp/dt=4.
Thanks in advance!
1 answer