Respuesta :
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 [tex]\bf \begin{array}{ccllll} quantity(x)&price(y)\\ -----&-----\\ 80&5000\\ 81&4970\\ 82&4940\\ 83&4910 \end{array}\\\\ -----------------------------\\\\[/tex]
[tex]\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[/tex]
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 [tex]\bf \cfrac{dp}{dx}=-60x+7100[/tex]
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 [tex]\bf \begin{array}{ccllll} quantity(x)&price(y)\\ -----&-----\\ 80&5000\\ 81&4970\\ 82&4940\\ 83&4910 \end{array}\\\\ -----------------------------\\\\[/tex]
[tex]\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[/tex]
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 [tex]\bf \cfrac{dp}{dx}=-60x+7100[/tex]
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
Price per person be lowered to maximize the profit for the agency is equals to $1150.
What is maximization ?
"Maximization is represented as the first derivative of profit function with respect to input function is equals to zero."
Formula used
'P' represents the profit
[tex]\frac{dP}{dx}=0[/tex]
Profit 'P' = Revenue 'R' - Cost 'C'
According to the question,
'x' represents the additional people
Number of people currently signed up for a tour =80
Total number of people = 80 +x
Price of a ticket per person = $5000
Ticket price = 5000-30x
Cost of plane seating 150 people = $250 000
Additional costs incidental fees of per person= $300
Additional costs incidental fees of x person= $300x
Price lowered = $30
Cost 'C'= 300x + 250,000
Revenue 'R' = (80 +x)(5000-30x)
Substitute the value in the formula we get,
Profit 'P' = [tex](80 +x)(5000-30x) -(300x + 250,000)[/tex]
= [tex]400,000 - 2400x + 5000x - 30x^{2} - 300x -250,000[/tex]
= [tex]-30x^{2} +2300x + 150,000[/tex]
Take first derivative we get,
[tex]\frac{dP}{dx} = -60x+2300 +0[/tex]
[tex]\frac{dP}{dx}=0[/tex]
Therefore,
[tex]-60x +2300 =0[/tex]
⇒[tex]x= \frac{2300}{60}[/tex]
⇒ [tex]x= 38.33[/tex]
Substitute the value of 'x' to get price per person be lowered
30x = 30 × 38.33
= $ 1149.9
≈$1150
Hence, price per person be lowered to maximize the profit for the agency is equals to $1150.
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