The revenue equation (in hundreds of millions of dollars) for barley production in a certain country is approximated by R(x) = 0.0613x​2 +​ 1.1584x + 2.4813 where x is in hundreds of millions of bushels. Find the marginal-revenue equation and use it to find the marginal revenue for the production of the given number of bushels.1) The marginal-revenue equation is R' (x) = _______2) Find the marginal revenue for the production of 200,000,000 bushels______3) Find the marginal revenue for the production of 650,000,000______ ​

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Answer:

The marginal-revenue equation is [tex]R'(x)=0.1226x+1.1584[/tex].

The marginal revenue for the production of 200,000,000 bushels is [tex]R'(x)=1.4036[/tex].

The marginal revenue for the production of 650,000,000 bushels is [tex]R'(x)=9.1274[/tex].

Step-by-step explanation:

The derivative [tex]R'(x)[/tex] of the revenue function is called the marginal revenue function and is the rate of change of revenue with respect to the number of units sold.

We know the revenue function, [tex]R(x) = 0.0613x^{2} +1.1584x+2.4813[/tex]. Therefore, the marginal revenue function is

[tex]R'(x)=\frac{d}{dx}R(x) = \frac{d}{dx}(0.0613x^{2} +1.1584x+2.4813)\\\\=\frac{d}{dx}\left(0.0613x^2\right)+\frac{d}{dx}\left(1.1584x\right)+\frac{d}{dx}\left(2.4813\right)\\\\R'(x)=0.1226x+1.1584[/tex]

To find the marginal revenue for the production of:

200,000,000 bushels we use x = 2, since x is in units of hundreds of million of dollars.

[tex]R'(x)=0.1226\cdot \:2+1.1584\\\\R'(x)=0.2452+1.1584\\\\R'(x)=1.4036[/tex]

650,000,000 bushels we use x = 65, since x is in units of hundreds of million of dollars.

[tex]R'(x)=0.1226\cdot \:65+1.1584\\\\R'(x)=7.969+1.1584\\\\R'(x)=9.1274[/tex]

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