Question

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______ ​

Answer 1

The marginal-revenue equation is
`R'(x)=0.1226x+1.1584`.

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

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

Explanation:

The derivative
`R'(x)` 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,
`R(x) = 0.0613x^(2) +1.1584x+2.4813`. Therefore, the marginal revenue function is

`R'(x)=(d)/(dx)R(x) = (d)/(dx)(0.0613x^(2) +1.1584x+2.4813)\\\\=(d)/(dx)\left(0.0613x^2\right)+(d)/(dx)\left(1.1584x\right)+(d)/(dx)\left(2.4813\right)\\\\R'(x)=0.1226x+1.1584`

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.

`R'(x)=0.1226\cdot \:2+1.1584\\\\R'(x)=0.2452+1.1584\\\\R'(x)=1.4036`

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

`R'(x)=0.1226\cdot \:65+1.1584\\\\R'(x)=7.969+1.1584\\\\R'(x)=9.1274`