For every 15 situps Ryan does, he does 10 push-ups. Create a table of equivalent ratios that show the relationship between the n
umber of sit-ups and the number of push-ups Ryan does.
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So hmm let's take a peek at the cost first
so, they chartered the plane for 150 folks with a fixed cost of 250,000
now, incidental fees are 300 per person, if we use the quantity "x", for how many folks, then if "x" persons are booked, then incidental fees are 300x
so, more than likely an insurance agency is charging them 300x for coverage
anyway, thus the cost C(x) = 250,000 + 300x
now, the Revenue R(x), is simple is jut price * quantity
well, the price, thus far we know is 5000 for 80 folks, but it can be lowered by 30 to get one more person, thus increasing profits
so... let's see what the price say y(x) is
so.. now we know y(x) = -30x+7400
now, Revenue is just price * quantity
the price y(x) is -30x+7400, the quantity is "x"
that simply means R(x) = -30x²+7400x
now, for the profit P(x)
the profit is simple, that is just incoming revenue minus costs, whatever is left, is profit
so P(x) = R(x) - C(x)
P(x) = (7400x - 30x²) - (250,000+300x)
P(x) = -30x² + 7100x - 250,000
now, where does it get maximized? namely, where's the maximum for P(x)?
well
and as you can see, if you zero out the derivative, there's only 1 critical point, run a first-derivative test on it, to see if its a maximum
so, they chartered the plane for 150 folks with a fixed cost of 250,000
now, incidental fees are 300 per person, if we use the quantity "x", for how many folks, then if "x" persons are booked, then incidental fees are 300x
so, more than likely an insurance agency is charging them 300x for coverage
anyway, thus the cost C(x) = 250,000 + 300x
now, the Revenue R(x), is simple is jut price * quantity
well, the price, thus far we know is 5000 for 80 folks, but it can be lowered by 30 to get one more person, thus increasing profits
so... let's see what the price say y(x) is
so.. now we know y(x) = -30x+7400
now, Revenue is just price * quantity
the price y(x) is -30x+7400, the quantity is "x"
that simply means R(x) = -30x²+7400x
now, for the profit P(x)
the profit is simple, that is just incoming revenue minus costs, whatever is left, is profit
so P(x) = R(x) - C(x)
P(x) = (7400x - 30x²) - (250,000+300x)
P(x) = -30x² + 7100x - 250,000
now, where does it get maximized? namely, where's the maximum for P(x)?
well
and as you can see, if you zero out the derivative, there's only 1 critical point, run a first-derivative test on it, to see if its a maximum