The number x of stereo speakers a retail chain is willing to sell per week at a price of p dollars is given by. Find the supply and instantaneous rate of change of supply when the price is 75 dollars?

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Answer:

Please find below the complete question:

The number  x  of stereo speakers a retail chain is willing to sell per week at a price of  p  dollars is given by  x  =  80 [tex]\sqrt{p+27}[/tex]  −  410.  Find the supply and instantaneous rate of change of supply when the price is  75  dollars.

Supply = 398

Instantaneous rate of change of supply [tex]=3.96[/tex]

Step-by-step explanation:

[tex]x=80\sqrt{p+27} -410[/tex]

Supply when price = $75

[tex]x=80\sqrt{75+27} -410 = x=80\sqrt{102} -410\\\\x = 80(10.1) -410= 808-410\\\\x = 398[/tex]

Supply = 398 stereo speakers

To get the instantaneous rate of supply, we differentiate the supply, x, with respect to price, p

[tex]\frac{dx}{dp} = \frac{d}{dp}(80\sqrt{p+27} -410)\\\\\frac{dx}{dp} = \frac{d}{dp}(80(p+27)^{1/2} -410)[/tex]

Using chain rule, we have

[tex]\frac{x}{y} \frac{dx}{dp}=\frac{d}{dp}( 80(p+27)^{1/2}) - \frac{d}{dp}(410) \\\\\frac{dx}{dp}=80(\frac{1}{2})(p+27)^{-1/2} - 0\\\\\frac{dx}{dp}=\frac{40}{\sqrt{p+27} } \\[/tex]

At p = $75

[tex]\frac{dx}{dp}=\frac{40}{\sqrt{75+27} }=\frac{40}{\sqrt{102} }\\\\\frac{dx}{dp}=\frac{40}{10.1} = 3.96\\\\\frac{dx}{dp}=3.96[/tex]

The question is not complete and the complete question says:

The number  x  of stereo speakers a retail chain is willing to sell per week at a price of  p  dollars is given by x=80√(p+27) − 410 Find the supply and instantaneous rate of change of supply when the price is  75  dollars

Answer:

A) Approximately 398 stereo speakers

B) Instantaneous rate of change = 3.961

Step-by-step explanation:

From the question, the number of speakers is given as;

x=80√(p+27) − 410

A) Thus at a price of $75,

The amount supplied will be;

x = 80√(75+27) − 410

x = 80√(102) - 410

x = (80 x 10.0995) - 410

x = 807.96 - 410 = 397.96 ≈ 398 stereo speakers

B) The instantaneous rate of change is given as;

dx/dp of the supply equation.

Thus, let's rearrange the supply equation first before differentiating.

So, x = 80√(p+27) − 410

x = 80[(p +27)^(1/2)] - 410

So, dx/dp = [(80 x 1/2) (p + 27)^(-1/2)]

dx/dp = 40/√(p+27)

So, at price of $75,

dx/dp = 40/√(75+27) = 40/√102

= 40/10.0995 = 3.961