Answer:
The demand reduces by $7.12 per month
Step-by-step explanation:
Given
[tex]p\to price[/tex]
[tex]x \to demand[/tex]
[tex]2x^2+5xp+50p^2=24800.[/tex]
[tex]p =10; \frac{dp}{dt} = 2[/tex]
Required
Determine the rate of change of demand
We have:
[tex]2x^2+5xp+50p^2=24800.[/tex]
Differentiate with respect to time
[tex]4x\frac{dx}{dt} + 5x\frac{dp}{dt} + 5p\frac{dx}{dt} + 100p\frac{dp}{dt} = 0[/tex]
Collect like terms
[tex]4x\frac{dx}{dt} + 5p\frac{dx}{dt} = -5x\frac{dp}{dt} - 100p\frac{dp}{dt}[/tex]
Factorize
[tex]\frac{dx}{dt}(4x + 5p) = -5(x + 20p)\frac{dp}{dt}[/tex]
Solve for dx/dt
[tex]\frac{dx}{dt} = -\frac{5(x + 20p)}{4x + 5p}\cdot \frac{dp}{dt}[/tex]
Given that: [tex]2x^2+5xp+50p^2=24800.[/tex] and [tex]p = 10[/tex]
Solve for x
[tex]2x^2 + 5x * 10 + 50 * 10^2 = 24800[/tex]
[tex]2x^2 + 50x + 5000 = 24800[/tex]
Equate to 0
[tex]2x^2 + 50x + 5000 - 24800 =0[/tex]
[tex]2x^2 + 50x -19800 =0[/tex]
Using a quadratic calculator, we have:
[tex]x \approx -113\ and\ x\approx88[/tex]
Demand must be greater than 0;
So: [tex]x=88[/tex]
So, we have: [tex]x=88[/tex]; [tex]p =10; \frac{dp}{dt} = 2[/tex]
The rate of change of demand is:
[tex]\frac{dx}{dt} = -\frac{5(88 + 20*10)}{4*88 + 5*10} * 2[/tex]
[tex]\frac{dx}{dt} = -\frac{5(288)}{402} * 2[/tex]
[tex]\frac{dx}{dt} = -\frac{2880}{402}[/tex]
[tex]\frac{dx}{dt} \approx -7.16[/tex]
This implies that the demand reduces by $7.12 per month
