Question

A travel agent currently has 80 people signed up for a tour. The price of a ticket is $5000 per person. The agency has chartered a plane seating 150 people at a cost of $250 000. Additional costs to the agency are incidental fees of $300 per person. For each $30 that the price is lowered, one new person will sign up. How much should the price per person be lowered to maximize the profit for the agency?

Answer 1

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  $\bf \begin{array}{ccllll} quantity(x)&price(y)\\ -----&-----\\ 80&5000\\ 81&4970\\ 82&4940\\ 83&4910 \end{array}\\\\ -----------------------------$

$\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 80}}\quad ,&{{ 5000}})\quad % (c,d) &({{ 83}}\quad ,&{{ 4910}}) \end{array} \\\quad \\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-90}{3}\implies -30 \\ \quad \\\\ % point-slope intercept y-{{ 5000}}={{ -30}}(x-{{ 80}})\implies y=-30x+2400+5000\\ \left.\qquad \right. \uparrow\\ \textit{point-slope form} \\\\\\ y=-30x+7400$

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 $\bf \cfrac{dp}{dx}=-60x+7100$

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