Market research has shown the price (p) and weekly sales (y(p) ) of a particular product are related by
the following differential equation:
(dy/dp) = (-1/2)* [y/ (p+3)]
If sales amount to 100 units when the price is $1 (i.e., y(1) = 100), what will the weekly sales be if the
price is raised to $6?
the following differential equation:
(dy/dp) = (-1/2)* [y/ (p+3)]
If sales amount to 100 units when the price is $1 (i.e., y(1) = 100), what will the weekly sales be if the
price is raised to $6?
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Separate variables to solve for y.
dy/y = -1/2 dp/(p + 3) ==> ln|y| = -(1/2) ln|p + 3| + C = ln(p + 3)^(-1/2) + C.
Exponentiate. We can assume that y > 0 since it's "sales".
y = K/√(p + 3)
where K = e^C.
y(1) = 100 ==> K/√(4) = 100 ==> K = 200.
So y(p) = 200/√(p + 3).
If p = 6, then y(6) = 200/3 units.
dy/y = -1/2 dp/(p + 3) ==> ln|y| = -(1/2) ln|p + 3| + C = ln(p + 3)^(-1/2) + C.
Exponentiate. We can assume that y > 0 since it's "sales".
y = K/√(p + 3)
where K = e^C.
y(1) = 100 ==> K/√(4) = 100 ==> K = 200.
So y(p) = 200/√(p + 3).
If p = 6, then y(6) = 200/3 units.