Home / Assignment Help / An 18 oz jar of peanut butter cost $2.19. A 48 oz jar of peanut butter sells for $4.39. Which size jar is the better buy?

An 18 oz jar of peanut butter cost $2.19. A 48 oz jar of peanut butter sells for $4.39. Which size jar is the better buy?

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

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