c)The price (in dollars) p and the quantity demanded q are related by the equation: p2+2q2=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.
i)Find A.
ii)Find dR/dt when q=20 and dp/dt=4.

i) A can be found by differentiating the equation p^2 + 2q^2 = 1100 with respect to q.

Let's differentiate both sides with respect to q:
2p * (dp/dq) + 4q = 0

Simplifying this equation, we get:
dp/dq = -2q / p

Now, we need to express dR/dt in terms of q. Given that dR/dt = A(dp/dt), we can substitute dp/dt with -2q / p:

dR/dt = A * (-2q / p)

ii) To find dR/dt when q = 20 and dp/dt = 4, we can substitute these values into the expression for dR/dt:

dR/dt = A * (-2 * 20 / p)

However, we don't have enough information to determine the value of p, so we can't calculate the exact value of dR/dt. In this case, we would need additional information, such as the value of p, in order to obtain a numerical answer.

To find A, we need to isolate dp/dt in the equation p^2 + 2q^2 = 1100.

p^2 + 2q^2 = 1100
Differentiate both sides with respect to t:
2p * dp/dt + 4q * dq/dt = 0

Now, solve for dp/dt:
dp/dt = (-4q * dq/dt) / (2p)

Since A is a function of just q, we can substitute dp/dt = 4 and q = 20 into the equation above to find dR/dt:

dR/dt = A * (dp/dt)
= A * 4

Therefore, to find A, we need to substitute the given values dp/dt = 4 and q = 20 into the equation dp/dt = (-4q * dq/dt) / (2p) and solve for A.

To find A, we need to express R in terms of q. We have the equation p^2 + 2q^2 = 1100, which means we can solve for p in terms of q.

Starting with p^2 + 2q^2 = 1100,
Rearranging the equation, we get:
p^2 = 1100 - 2q^2
Taking the square root of both sides, we have:
p = √(1100 - 2q^2)

Now, we can express R in terms of q:
R = pq

To find A, we need to differentiate R with respect to t. However, since we only have q and p in the equation, we need to use the chain rule. Recall the equation provided:

dR/dt = A(dp/dt)

Using the chain rule, we can express dR/dt in terms of q and dp/dt:

dR/dt = ∂R/∂q * dq/dt + ∂R/∂p * dp/dt

Since we have R = pq, we can find the partial derivatives:
∂R/∂q = p
∂R/∂p = q

Substituting these values into the chain rule equation, we have:

dR/dt = p * dq/dt + q * dp/dt

Now, we need to express p and q in terms of each other using the equation p^2 + 2q^2 = 1100:

From the earlier expression, we have:
p = √(1100 - 2q^2)

Substituting this into the chain rule equation:

dR/dt = (√(1100 - 2q^2)) * dq/dt + q * dp/dt

This equation is in terms of p and q, but we want to express it in terms of q only. To do that, we need to find dq/dt and dp/dt.

Given that dp/dt = 4 (as stated in the question), we can plug in q = 20 and dp/dt = 4 into the equation:

dq/dt = (∂q/∂t) = (dp/dt) / (∂p/∂q)

To find (∂p/∂q), we differentiate the expression p = √(1100 - 2q^2) with respect to q:

∂p/∂q = (-2q) / (2√(1100 - 2q^2))

Plugging in q = 20, (∂p/∂q) = (-2 * 20) / (2√(1100 - 2(20)^2))

Simplifying, we get (∂p/∂q) = -40 / (2√(1100 - 800)) = -20 / (√300) = -20 / (10√3) = -2 / √3

Now we can find dq/dt:

dq/dt = (dp/dt) / (∂p/∂q) = 4 / (-2 / √3) = -4 * √3 / 2 = -2√3

Finally, we can find dR/dt when q = 20 and dp/dt = 4:

dR/dt = (√(1100 - 2q^2)) * dq/dt + q * dp/dt
Substituting q = 20 and dq/dt = -2√3, we get:
dR/dt = (√(1100 - 2(20)^2)) * (-2√3) + 20 * 4

Simplifying further:
dR/dt = (√(1100 - 800)) * (-2√3) + 80
dR/dt = (√300) * (-2√3) + 80
dR/dt = -√(300) * 2√3 + 80
dR/dt = -2√(3*100) + 80
dR/dt = -2 * 10√3 + 80
dR/dt = -20√3 + 80

Therefore, when q = 20 and dp/dt = 4, dR/dt is equal to -20√3 + 80.

p^2 + 2 q^2 = 1100

2 p dp/dt + 4 q dq/dt = 0
dp/dt = -(q/p) dq/dt

R = p q price times quantity
dR/dt = p dq/dt + q dp/dt = A dp/dt

so A is just q

when q = 20
p^2 + 2(400) = 1100
p^2 = 300
p = 10 sqrt 3

dR/dt = q dp/dt = 20*4 = 80