Step-by-Step Solution

Topic: Maxima & Minima

(I) FIND F AT V = 40 km/h

Given function: $$ F = \frac{V^2}{500} – \frac{V}{4} + 14 $$

Substitute V = 40:

$$ F = \frac{40^2}{500} – \frac{40}{4} + 14 $$

$$ F = \frac{1600}{500} – 10 + 14 $$

$$ F = 3.2 – 10 + 14 $$

$$ F = 7.2 $$

(II) FIND dF/dV

Differentiate F with respect to V:

$$ \frac{dF}{dV} = \frac{d}{dV}\left(\frac{V^2}{500}\right) – \frac{d}{dV}\left(\frac{V}{4}\right) + \frac{d}{dV}(14) $$

$$ \frac{dF}{dV} = \frac{2V}{500} – \frac{1}{4} + 0 $$

$$ \frac{dF}{dV} = \frac{V}{250} – \frac{1}{4} $$

(III)(A) MINIMIZE FUEL CONSUMPTION

For minimum value, set dF/dV = 0 (Critical Point):

$$ \frac{V}{250} – \frac{1}{4} = 0 $$

$$ \frac{V}{250} = \frac{1}{4} $$

$$ V = \frac{250}{4} = 62.5 $$

Verification (Second Derivative Test):

$$ \frac{d^2F}{dV^2} = \frac{1}{250} $$

Since the second derivative is positive (>0), V = 62.5 corresponds to a Minimum.

(III)(B) FUEL FOR 600 KM

Step 1: Find Speed V

Given dF/dV = -0.01.

$$ \frac{V}{250} – \frac{1}{4} = -0.01 $$

$$ \frac{V}{250} – 0.25 = -0.01 $$

$$ \frac{V}{250} = 0.24 $$

$$ V = 250 \times 0.24 = 60 \text{ km/h} $$

Step 2: Find Consumption Rate (F) at V = 60

$$ F = \frac{60^2}{500} – \frac{60}{4} + 14 $$

$$ F = \frac{3600}{500} – 15 + 14 $$

$$ F = 7.2 – 1 = 6.2 \text{ l/100 km} $$

Step 3: Calculate Total Fuel for 600 km

Rate is 6.2 litres for every 100 km.

Total Fuel = (6.2 / 100) × 600

Total Fuel = 6.2 × 6 = 37.2 litres

⚠️ Silly Mistake Audit: Unit Interpretation

Check the Definition of F!
A common mistake in Part (iii)(b) is assuming F is the “total fuel” or “fuel per km”.
The question defines F as litres per 100 km.
If you calculate F = 6.2 and then just multiply by 600 (thinking it’s per km), you get 3720 litres, which is wrong. Always remember to divide by 100 if the unit is “l/100 km”.