# Can somehow help justify if my solutions to the problem below is correct and guide me on (b)?

## Oct 19, 2016

The customer must travel $6 \frac{2}{3}$ Km to reduce average cost at $80$cents/kilometer.

#### Explanation:

b) let x be the asked kilometer for average cost at 80 cents per kilometer. Then $x \cdot 0.80 = 2 + \left(0.50 \cdot x\right) \mathmr{and} 0.80 x - 0.50 x = 2 \mathmr{and} 0.30 x = 2 \mathmr{and} x = \frac{2}{3} \cdot 10 = \frac{20}{3} = 6 \frac{2}{3} k m$[Ans]

Oct 19, 2016

For a distance of $6 \frac{2}{3}$ km the average cost will be $80 c$

#### Explanation:

$x$ is the number of Km travelled.

The total cost of travelling $x$ km is calculated as:

50c" per km" + $2" "larr 50c =$0.50

$\downarrow \text{ "darr " } \downarrow$

$0.5 \times \text{x" } + 2 = 0.5 x + 2$

To find the average cost per Km:
divide the total cost by the number of Km travelled.

Ave cost = $f \left(x\right) = \frac{0.5 x + 2}{x}$

The average cost for $x$ km should be 80c rarr $0.8 $\frac{0.5 x + 2}{x} = 0.8 \text{ } \leftarrow \times x$$0.5 x + 2 = 0.8 x \text{ } \leftarrow$move x-terms to the right $2 = 0.8 x - 0.5 x$$2 = 0.3 x \text{ } \leftarrow$divide by 0.3 $\frac{2}{0.3} = x$$\frac{20}{3} = x$$x = 6 \frac{2}{3}$For a distance of $6 \frac{2}{3}$km the average cost will be $80 c\$