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Suppose a granola bar company estimates that its monthly cost is ๐ถ(๐‘ฅ)=500๐‘ฅ2+400๐‘ฅC(x)=500x 2 +400x and its monthly revenue is ๐‘…(๐‘ฅ)=โˆ’0.6๐‘ฅ3+800๐‘ฅ2โˆ’300๐‘ฅ+600R(x)=โˆ’0.6x 3 +800x 2 โˆ’300x+600, where x is in thousands of granola bars sold. The profit is the difference between the revenue and the cost.What is the profit function, P(x)?A.๐‘ƒ(๐‘ฅ)=โˆ’0.6๐‘ฅ3+1300๐‘ฅ2+100๐‘ฅ+600P(x)=โˆ’0.6x 3 +1300x 2 +100x+600B.๐‘ƒ(๐‘ฅ)=0.6๐‘ฅ3+300๐‘ฅ2โˆ’700๐‘ฅ+600P(x)=0.6x 3 +300x 2 โˆ’700x+600C.๐‘ƒ(๐‘ฅ)=โˆ’0.6๐‘ฅ3+300๐‘ฅ2โˆ’700๐‘ฅ+600P(x)=โˆ’0.6x 3 +300x 2 โˆ’700x+600D.๐‘ƒ(๐‘ฅ)=0.6๐‘ฅ3โˆ’300๐‘ฅ2+700๐‘ฅโˆ’600P(x)=0.6x 3 โˆ’300x 2 +700xโˆ’600SUBMITarrow_backPREVIOUS

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Suppose a granola bar company estimates that its monthly cost is ๐ถ(๐‘ฅ)=500๐‘ฅ2+400๐‘ฅC(x)=500x 2 +400x and its monthly revenue is ๐‘…(๐‘ฅ)=โˆ’0.6๐‘ฅ3+800๐‘ฅ2โˆ’300๐‘ฅ+600R(x)=โˆ’0.6x 3 +800x 2 โˆ’300x+600, where x is in thousands of granola bars sold. The profit is the difference between the revenue and the cost.What is the profit function, P(x)?A.๐‘ƒ(๐‘ฅ)=โˆ’0.6๐‘ฅ3+1300๐‘ฅ2+100๐‘ฅ+600P(x)=โˆ’0.6x 3 +1300x 2 +100x+600B.๐‘ƒ(๐‘ฅ)=0.6๐‘ฅ3+300๐‘ฅ2โˆ’700๐‘ฅ+600P(x)=0.6x 3 +300x 2 โˆ’700x+600C.๐‘ƒ(๐‘ฅ)=โˆ’0.6๐‘ฅ3+300๐‘ฅ2โˆ’700๐‘ฅ+600P(x)=โˆ’0.6x 3 +300x 2 โˆ’700x+600D.๐‘ƒ(๐‘ฅ)=0.6๐‘ฅ3โˆ’300๐‘ฅ2+700๐‘ฅโˆ’600P(x)=0.6x 3 โˆ’300x 2 +700xโˆ’600SUBMITarrow_backPREVIOUS

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Solution

The profit function, P(x), is found by subtracting the cost function, C(x), from the revenue function, R(x). This means we have:

P(x) = R(x) - C(x)

Substituting the given functions into this equation gives:

P(x) = (-0.6x^3 + 800x^2 - 300x + 600) - (500x^2 + 400x)

Simplify this by combining like terms:

P(x) = -0.6x^3 + (800x^2 - 500x^2) + (-300x - 400x) + 600 P(x) = -0.6x^3 + 300x^2 - 700x + 600

So, the correct answer is C. ๐‘ƒ(๐‘ฅ)=โˆ’0.6๐‘ฅ3+300๐‘ฅ2โˆ’700๐‘ฅ+600.

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