Question

suppose that the price p, in dollars, and the number of sales, x, of a certain item are related by 4p + 5x + 2px = 50. if p and x are both functions of time, measured in days, find the rate at which x is changing when x = 2, p = 5, and \\( \frac{dp}{dt} = 1.8 \\).
the rate at which x is changing is \\( \square \\)
(round to the nearest hundredth as n

sale(s)
day(s) per sale
day(s)
sale(s) per day

Explanation:

Step1: Differentiate the equation with respect to \(t\)

Differentiate \(4p + 5x+2px = 50\) term - by - term using the sum rule \((u + v+w)'=u'+v'+w'\) and the product rule \((uv)' = u'v+uv'\).
The derivative of \(4p\) with respect to \(t\) is \(4\frac{dp}{dt}\), the derivative of \(5x\) with respect to \(t\) is \(5\frac{dx}{dt}\), and for the term \(2px\), using the product rule \((uv)^\prime=u^\prime v + uv^\prime\) (where \(u = 2p\) and \(v=x\)), its derivative is \(2\frac{dp}{dt}x+2p\frac{dx}{dt}\). The derivative of the constant \(50\) is \(0\).
So, \(4\frac{dp}{dt}+5\frac{dx}{dt}+2x\frac{dp}{dt}+2p\frac{dx}{dt}=0\).

Step2: Substitute the given values \(x = 2\), \(p = 5\), and \(\frac{dp}{dt}=1.8\)

Substitute into the equation \(4\frac{dp}{dt}+5\frac{dx}{dt}+2x\frac{dp}{dt}+2p\frac{dx}{dt}=0\):

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Answer:

\(-0.96\) sale(s) per day