Question

"A contributing factor to an airplane's duel consumption is the bypass ratio of the engine system. The bypass ratio is the amount of air passing around the engine core relative to the amount of air passing through the core. An airplane manufacturer is designing a new airplane and wants to determine the bypass ratio for the airplane's engine system. The airplane will fly 3,800 hours per year and will average 560 miles per hour. The amount of fuel that the airplane consumes can be expressed as:
z = 0.0469 - (7.94*10^-4) * y
for 4 < y < 12
where y is the bypass ratio (a unitless number) and z is the number of gallons of fuel consumed per mile flown by the airplane. The cost of fuel remains constant at $4.33 per gallon.
The initial cost of the engine system as a function of the bypass ratio is $267,000 + $2,600y^2.
The engine system will be used for 13 years. At the end of 13 years, the salvage value of the engine system as a function of bypass ratio is $7,000y. The airplane manufacturer wants to minimize the annual equivalent cost (AEC) of the engine system (which includes the initial cost, the annual cost of fuel, and the salvage value). The manufacturer's MARR is 13.1%. What is the optimal bypass ratio rounded to the nearest tenth that minimizes the AEC of the engine system?
(The optimal answer for the bypass ratio is between 4 to 12, but it should not be necessary to consider that constraint in your calculations.)"

Answer 1

Initial cost of engine (IE) = 267,000 + 2,600y²

Salvage value (SV) = 7000y

Amount of fuel consumed in a year = 3800 x 560 x 0.0469 - (7.94 x 10^-4) y

Cost of fuel per year (AF)= 4.33 x 3800 x 560 x [0.0469 - (7.94 x 10^-4) y]

It is required to minimize IE - SV + AF => 267000 + 2,600y² - 7000y + 9214240 [0.0469 - (7.94 x 10^-4 ) y]

= 267000 + 2600y² - 7000y + 432147.85 - 7316.10y

= 2600y² - 14316.10y + 699147.85

Taking derivative of the expression and setting it equal to zero will minimize the value of the expression at desired y.

5200y - 14316.10 = 0

y = 2.75