Question

The fuel flowing to an engine of a small aircraft can be modelled by a quadratic

equation over a limited range of speeds using the relation f = 0.0048v2
- 0.96v + 64
where f represents the flow of fuel, in litres per hour, and v represents speed, in kilometres per hour. Determine the speed that minimizes fuel flow.

Answer 1

Answer: So, the speed that minimizes fuel flow is 400 km/h.

Explanation:

To determine the speed that minimizes fuel flow, we need to find the vertex of the parabola represented by the equation f = 0.0048v^2 - 0.96v + 64. The vertex of a parabola in the form of y = a(x - h)^2 + k is (h, k).

To find h, we can use the formula h = -b / (2a) = -(-0.96) / (2 * 0.0048) = 400

To find k, we can substitute h and v = 400 into the original equation:

f = 0.0048(400)^2 - 0.96(400) + 64 = 64

So, the speed that minimizes fuel flow is 400 km/h.