The revenue equation (in hundreds of millions of dollars ) for barley production

Question

The revenue equation (in hundreds of millions of dollars ) for barley production in a certain country is approximate by R(x)=0.0633x^2+1.3988x+2.1475 where x is in hundreds of millions of nushels. Find the marginal revenue equation and use it to find the marginal revenue for the production of the gien number of bushels.

a.0 The marginal revenue equation is R ' (x)=

b. Find the marginal revenue for the production of $600,000,000 bushels

c.Find the marginal revenue for the production of $850,000,000 bushels

Explanation / Answer

a.Marginal Revenue equation=>2*0.0633x+1.3988=R'(x) (Obtained by differentiating the Revenue Equation)
R'(x)=0.1266x+1.3988

b. Production=R(x)=6 because R(x) is expressed in terms of hundred of millions of dollar, which in this case is 6 hundred million dollars
R(x)=0.0633x^2+1.3988x+2.1475=6
Solving the above quadratic equation, we get x1=2.4765 & x2=-24.57(Rejected because x is the number of hundreds of millions of nushels and hence can't be -ve)
therefore, x=2.4765 hundreds of millions of nushels
Marginal Revenue= 0.1266x+1.3988
putting the value of x=2.4765, we get R'(x)=1.71232 hundreds of millions of dollar

c.Production=R(x)=8.5 because R(x) is expressed in terms of hundred of millions of dollar, which in this case is 8.5 hundred million dollars
R(x)=0.0633x^2+1.3988x+2.1475=8.5
Solving the above quadratic equation, we get x1=3.8652 & x2=-25.96(Rejected because x is the number of hundreds of millions of nushels and hence can't be -ve)
therefore, x=3.8652 hundreds of millions of nushels
Marginal Revenue= 0.1266x+1.3988
putting the value of x=3.8652, we get R'(x)=1.888 hundreds of millions of dollar

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