suppose that the price p, in dollars, and the number of sales, x, of a certain item are related by 4p + 5x +…

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 \\).\nthe rate at which x is changing is \\( \\square \\)\n(round to the nearest hundredth as n\n\n sale(s)\n day(s) per sale\n day(s)\n sale(s) per day

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 \\).\nthe rate at which x is changing is \\( \\square \\)\n(round to the nearest hundredth as n\n\n sale(s)\n day(s) per sale\n day(s)\n sale(s) per day

Answer

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): [ \begin{align*} 4\times1.8+5\frac{dx}{dt}+2\times2\times1.8+2\times5\times\frac{dx}{dt}&=0\ 7.2 + 5\frac{dx}{dt}+7.2+10\frac{dx}{dt}&=0\ 14.4+(5 + 10)\frac{dx}{dt}&=0\ 15\frac{dx}{dt}&=- 14.4\ \frac{dx}{dt}&=\frac{-14.4}{15} \end{align*} ]

Answer:

(-0.96) sale(s) per day