# Answer to Question #17245 in Calculus for hsd

Question #17245

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.

A=

Find dR/dt when q=20 and dp/dt=3.

dR/dt=

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.

A=

Find dR/dt when q=20 and dp/dt=3.

dR/dt=

Expert's answer

differentiate 2*p*dp+4*q*dq=0

p*dp/dt=-2*q*dq/dt

p=sqrt(1100-2q^2)

dp/dt=-2*q/sqrt(1100-2q^2) *dq/qt

A=A( -2*q/sqrt(1100-2q^2) *dq/qt )

q=20 dp/dt=3p=sqrt(1100-2q^2)=sqrt(1100-2*400)=sqrt(300)=10sqrt(3)

p*3=-40*dq/dt

dq/dt=-3/40 *10sqrt(3)=-3sqrt(3)/4

dR/dt=A( -2*q/sqrt(1100-2q^2) *dq/qt )=A(-40/(10sqrt(3))*-3sqrt(3)/4)=

=A(1/(sqrt(3)) *3sqrt(3))=A(3)

p*dp/dt=-2*q*dq/dt

p=sqrt(1100-2q^2)

dp/dt=-2*q/sqrt(1100-2q^2) *dq/qt

A=A( -2*q/sqrt(1100-2q^2) *dq/qt )

q=20 dp/dt=3p=sqrt(1100-2q^2)=sqrt(1100-2*400)=sqrt(300)=10sqrt(3)

p*3=-40*dq/dt

dq/dt=-3/40 *10sqrt(3)=-3sqrt(3)/4

dR/dt=A( -2*q/sqrt(1100-2q^2) *dq/qt )=A(-40/(10sqrt(3))*-3sqrt(3)/4)=

=A(1/(sqrt(3)) *3sqrt(3))=A(3)

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