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) 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) To find the rate of change of the demand, we need to find dx/dt (the rate of change of demand).
First, we have the equation 2x^2 + 6xp + 50p^2 = 10600.
To find dx/dt, we need to differentiate both sides of the equation with respect to t (time).
Differentiating each term with respect to t, we get:
d(2x^2)/dt + d(6xp)/dt + d(50p^2)/dt = d(10600)/dt
Simplifying, we have:
2(2x)(dx/dt) + 6x(dp/dt) + 2(50p)(dp/dt) = 0
Now, we can substitute the given values: p = $30 (price), and dp/dt = $4 (rate of change of price).
Substituting these values and solving for dx/dt, we have:
2(2x)(dx/dt) + 6x(4) + 2(50p)(4) = 0
4x(dx/dt) + 24x + 400p = 0
4x(dx/dt) = -24x - 400p
dx/dt = (-24x - 400p) / (4x)
Now, substitute p = $30, x = ? (demand), and dp/dt = $4 into the equation to find the rate of change of the demand.
2) a) To find dR/dt, we need to first find the derivative of R with respect to t (time).
Given that R = pq (revenue), where p is the price and q is the quantity demanded, we can rewrite the equation: p^2 + 2q^2 = 1100.
Differentiating each term with respect to t, we get:
d(p^2)/dt + d(2q^2)/dt = d(1100)/dt
Simplifying, we have:
2p(dp/dt) + 4q(dq/dt) = 0
Now, we can substitute the given value: q = 20 (quantity demanded), and dp/dt = 4 (rate of change of price).
Substituting these values into the equation, we have:
2p(4) + 4(20)(dq/dt) = 0
8p + 80(dq/dt) = 0
Since A is a function of just q, we can isolate dq/dt by dividing the equation by 80:
dq/dt = -8p/80
dq/dt = -p/10
b) Now, we need to find dR/dt when q = 20 and dp/dt = 4.
Since R = pq, we differentiate R with respect to t (time):
dR/dt = p(dq/dt) + q(dp/dt)
Substituting q = 20, dq/dt = -p/10, and dp/dt = 4, we have:
dR/dt = p(-p/10) + 20(4)
dR/dt = -p^2/10 + 80
Now, substitute p = ?, into the equation to find dR/dt.
1) To find the rate of change of demand, we need to find dx/dt, the derivative of demand with respect to time.
Given: 2x^2 + 6xp + 50p^2 = 10600.
First, let's differentiate both sides of the equation with respect to time (t):
d(2x^2 + 6xp + 50p^2)/dt = d(10600)/dt.
Next, we can apply the chain rule to differentiate each term on the left-hand side:
2*(d(x^2)/dt) + 6x*(dp/dt) + 6p*(dx/dt) + 100p*(dp/dt) = 0.
Since we are asked to find dx/dt, we can solve for it:
6x*(dp/dt) + 100p*(dp/dt) = -2*(d(x^2)/dt).
We know that dp/dt = 4 (price is increasing at a rate of 4 dollars per month), and we are given that the price is 30 dollars:
6x*(4) + 100(30)*(4) = -2*(d(x^2)/dt).
Now let's solve for dx/dt:
24x + 12000 = -2*(d(x^2)/dt).
Since we want to find the rate of change of demand when the price is 30 dollars, we substitute p = 30 into the equation:
24x + 12000 = -2*(d(x^2)/dt).
24x + 12000 = -2*(2x*(dx/dt)), as d(x^2)/dt = 2x*(dx/dt).
Substituting the given price:
24x + 12000 = -2*(2*(30)*(dx/dt)).
24x + 12000 = -120*(dx/dt).
Now we can solve for dx/dt:
dx/dt = -(24x + 12000) / 120.
This is the rate of change of the demand.
2)
a) Given: p^2 + 2q^2 = 1100.
To find dR/dt, we need to differentiate the revenue equation, R = p*q, with respect to time:
dR/dt = dp/dt * q + p * dq/dt.
b) Given: q = 20 and dp/dt = 4.
Substituting the given values into the equation for dR/dt:
dR/dt = (4) * 20 + (p) * (dq/dt).
Since A is a function of just q, A = q^2. Therefore, dq/dt = d(A)/dt.
Substituting q = 20 and A = q^2 = 20^2 = 400 into the equation for dq/dt:
dq/dt = d(400)/dt = 0.
Now we can solve for dR/dt:
dR/dt = (4) * 20 + (p) * (0).
dR/dt = 80 + 0.
dR/dt = 80.
Therefore, when q = 20 and dp/dt = 4, the rate of change of revenue is 80 dollars per unit time.