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QuestionAn airplane flies 3000 miles at 550 mph. Cost per passenger is C(x)=150+x25+32,000xC(x)=150+\frac{x}{25}+\frac{32,000}{x}. Find C(550)C(550).

Studdy Solution

STEP 1

Assumptions1. The airspeed of the airplane is550 miles per hour. The cost per passenger is given by the function C(x)=150+x25+32,000xC(x)=150+\frac{x}{25}+\frac{32,000}{x}3. The variable xx represents the ground speed, which is the airspeed ± wind speed4. In quiescent conditions, there is no wind, so the ground speed is equal to the airspeed5. We need to find the cost per passenger for quiescent conditions

STEP 2

For quiescent conditions, the ground speed xx is equal to the airspeed. So, we substitute the airspeed into the cost function.
C(550)=150+55025+32,000550C(550)=150+\frac{550}{25}+\frac{32,000}{550}

STEP 3

implify the equation by performing the operations in the correct order according to the order of operations (PEMDAS/BODMAS).
First, perform the division in the second and third terms.
C(550)=150+22+32,000550C(550)=150+22+\frac{32,000}{550}

STEP 4

Continue simplifying the equation by performing the division in the third term.
C(550)=150+22+58.1818C(550)=150+22+58.1818

STEP 5

Finally, add up all the terms to find the cost per passenger for quiescent conditions.
C(550)=150+22+58.1818=230.1818C(550)=150+22+58.1818=230.1818

STEP 6

Round the cost to the nearest cent as needed.
C(550)=$230.18C(550)=\$230.18The cost per passenger for quiescent conditions is $230.18.

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