Question

The marginal revenue for a new calculator is given by 20,000 MR = 60,000 - (10 + x)2 where x represents hundreds of calculators and revenue is in dollars. Find the total revenue function, R(x), for these calculators R(X) - 1940000x 2000023 3 + 200000.x? Your answer cannot be understood or graded. More Information Need Help? Read it Talk to a Tutor Viewing Saved Work Revert to Last Response

Answer 1

First question's solution

Given, MR= 60,000 - 20,000/(10+x)²

MR = d(TR)/dx

TR = $\int$ MR dx = $\int$ {(60,000 - 20,000/(10+x)² }dx = $\int$ 60,000dx - $\int$ 20,000(10+x)^-2 dx

$\int$60,000dx = 60,000x +C_{​​​​​​1} Formula used are $\int$ kdz = kz +C where k is a constant value

Solving $\int$ 20,000*(10+x)^-2 dx . Let (10+x) = u

d(10+x) = du or d(10) + dx = du or dx = du . Substituting u for x, we get

$\int$20,000u^{-​​​​​​2}​​ du =  20,000 u^{​​​​​​-2+1} / (-2+1) + C_{​​​​​​2} = -20,000 u^{​​​​​-1}​= 20,000/u

$\int$(x)ⁿ dx = (x)ⁿ⁺¹ / (n+1) + C, where n$\neq$ -1 and a is constant. C_{​​​​​​} is integration constant

Using the value u = (10+x) we get

TR = $\int$ 60,000dx - $\int$ 20,000(10+x)^-2 dx = 60,000x + C_{​​​​​​1} + 20,000/(a+x) + C_{​​​​​​2}​​​

TR = R(x) = 60,000x + 1000/(10+x) + C , where C= C_{​​​​​​1} + C_{​​​​​​2}

Second question's solution

We denote average cost by AC instead of C bar . given

AC'(x) = (1/4) - (100/x^{​​​​​​2})

AC = $\int$ AC'(x) dx = $\int$ [(1/4) - (100/x^{​​​​​​2}) ]dx = $\int$ (1/4)dx - $\int$ (100x^{​​​​-2} ) dx

Using the formula

$\int$(x)ⁿ dx = (x)ⁿ⁺¹ / (n+1) + C, where n$\neq$ -1 and a is constant. C_{​​​​​​} is integration constant

$\int$kdz = kz +C where k is a constant value

AC(x) = x/4 + C_{​​​​​​1} - 100 x^{​​​​​​-2+1} / (-2+1) +C_{​​​​​​2} = (x/4) +100x^-1

AC(x) = (x/4) + (100/x)

At Q = 60, AC (20) = (60/4) + (100/60) = $50/3 = $16.667