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

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

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Answer 1 · 2016-10-19T11:40:50

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

Explanation:

b) let x be the asked kilometer for average cost at 80 cents per kilometer. Then $x \cdot 0.80 = 2 + (0.50 \cdot x) or 0.80 x - 0.50 x = 2 or 0.30 x = 2 or x = \frac{2}{3} \cdot 10 = \frac{20}{3} = 6 \frac{2}{3} k m$ $x*0.80 = 2+(0.50*x) or 0.80x-0.50x=2 or0.30x=2 or x=2/3*10 =20/3=6 2/3 km$ [Ans]

Answer 2 · 2016-10-19T13:29:40

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

Explanation:

$x$ $x$ is the number of Km travelled.

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

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

$↓ ↓ ↓$ $darr" "darr " "darr$

$0.5 \times x + 2 = 0.5 x + 2$ $0.5xx ""x" "+2 = 0.5x+2$

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

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

The average cost for $x$ $x$ km should be $80 c \to $ 0.8$ $80c rarr $0.8$

$\frac{0.5 x + 2}{x} = 0.8 \leftarrow \times x$ $(0.5x+2)/x = 0.8" "larr xx x$

$0.5 x + 2 = 0.8 x \leftarrow$ $0.5x +2 = 0.8x" "larr$ move x-terms to the right

$2 = 0.8 x - 0.5 x$ $2 = 0.8x-0.5x$

$2 = 0.3 x \leftarrow$ $2 = 0.3x" "larr$ divide by 0.3

$\frac{2}{0.3} = x$ $2/0.3 = x$

$\frac{20}{3} = x$ $20/3 = x$

$x = 6 \frac{2}{3}$ $x = 6 2/3$

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

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

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