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QuestionFind the unit price pp that maximizes revenue r(p)=−9p2+27,000pr(p) = -9p^2 + 27,000p and determine the maximum revenue.

Studdy Solution

STEP 1

Assumptions1. The revenue function is given by r(p)=−9p+27,000pr(p) = -9p^+27,000p . We need to find the unit price pp that maximizes the revenue rr
3. We also need to find the maximum revenue

STEP 2

The given function is a quadratic function. The maximum or minimum of a quadratic function ax2+bx+cax^2 + bx + c is given by −b/2a-b/2a. Here, a=−9a = -9 and b=27,000b =27,000.
pmax=−b2ap_{max} = -\frac{b}{2a}

STEP 3

Substitute the values of aa and bb into the formula to find the unit price that maximizes the revenue.
pmax=−27,0002×−9p_{max} = -\frac{27,000}{2 \times -9}

STEP 4

Calculate the value of pmaxp_{max}.
pmax=−27,000−18=1,500p_{max} = -\frac{27,000}{-18} =1,500

STEP 5

Now that we have the unit price that maximizes the revenue, we can substitute this value into the revenue function to find the maximum revenue.
rmax=r(pmax)=−9pmax2+27,000pmaxr_{max} = r(p_{max}) = -9p_{max}^2+27,000p_{max}

STEP 6

Substitute the value of pmaxp_{max} into the revenue function.
rmax=−9(1,500)2+27,000(1,500)r_{max} = -9(1,500)^2+27,000(1,500)

STEP 7

Calculate the value of rmaxr_{max}.
rmax=−9(2,250,000)+40,500,000=20,250,000r_{max} = -9(2,250,000)+40,500,000 =20,250,000The unit price that should be established to maximize revenue is 1,500andthemaximumrevenueis1,500 and the maximum revenue is 20,250,000.

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