Respuesta :
Plugging in p = 10 to solve for x:
x^2-50x-600 = (x-60)(x+10) = 0 => x = 60
x^2 - 5xp + 50p^2 = 3100
Differentiate with respect to time,
2xx' - 50x'p - 50xp' + 100pp' = 0
240 - 100(10) - 3000p' + 1000p' = 0
p' = -.38
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x^2-50x-600 = (x-60)(x+10) = 0 => x = 60
x^2 - 5xp + 50p^2 = 3100
Differentiate with respect to time,
2xx' - 50x'p - 50xp' + 100pp' = 0
240 - 100(10) - 3000p' + 1000p' = 0
p' = -.38
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
Using implicit differentiation, it is found that the rate of change of demand is of 21.64.
The equation that describes the relationship between price p and demand x is given by:
[tex]2x^2 - 10xp + 50p^2 = 6200[/tex]
In this problem, we have that the price is [tex]p = 10[/tex], thus, the demand is found solving the equation:
[tex]2x^2 - 100x + 5000 = 6200[/tex]
[tex]2x^2 - 100x - 1200 = 0[/tex]
[tex]x^2 - 10x - 120 = 0[/tex]
Which is a quadratic equation with [tex]a = 1, b = 10, c = -120[/tex]. Then:
[tex]\Delta = 10^2 - 4(1)(-120) = 580[/tex]
[tex]x_{1} = \frac{-10 + \sqrt{580}}{2} = 7.04[/tex]
[tex]x_{2} = \frac{-10 - \sqrt{580}}{2} = -17.04[/tex]
Since the demand is positive, we use [tex]x = 7.04[/tex].
The rate of change, applying implicit differentiation, is given by:
[tex]2x\frac{dx}{dt} - 10p\frac{dx}{dt} - 10x\frac{dp}{dt} + 100p\frac{dp}{dt} = 0[/tex]
We have that [tex]\frac{dp}{dt} = 2[/tex], thus:
[tex]2(7.04)\frac{dx}{dt} - 100\frac{dx}{dt} - 10(7.04)(2) + 100(10)(2) = 0[/tex]
[tex]\frac{dx}{dt}= \frac{10(7.04)(2) - 100(10)(2)}{2(7.04) - 100}[/tex]
[tex]\frac{dx}{dt} = 21.64[/tex]
The rate of change of demand is of 21.64.
A similar problem is given at https://brainly.com/question/25081524
