Question 414458
Let {{{x}}} = crossings per year it takes, on average, for two six-months passes to be the
most economical choice.
Let {{{C}}} = cost per year to make {{{x}}} crossings
Assume a constant number of trips per month
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Without buying any passes:
{{{C[1] = 3x}}}
Buying 2 6-month passes:
{{{C[2] = .5x + 2*15}}}
Buying a 1-year pass:
{{{C[3] = 150}}}
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First I'll set {{{C[1] = C[2]}}} to find intersection
{{{3x = .5x + 2*15}}}
{{{3x = .5x + 30}}}
{{{2.5x = 30}}}
{{{x = 12}}}
12 crossings per year make {{{C[1] = C[2]}}}
One more crossing, {{{x = 13}}}, make
{{{C[1] = 3*13}}}
{{{C[1] = 39}}}
and
{{{C[2] = .5*13 + 30}}}
{{{C[2] = 36.5}}}
So, {{{C[2]}}} becomes more economical on the 13th crossing by $2.50
and it still cost less than the 1-year pass for $150