Answer:
A) (20-(x-10)) (x-6) = -x^2 +36x-180
B)In order to maximize the weekly profit the feeder price should be $80 and the weekly profit will be $144
Explanation
A)Let's consider they priced each feeder dollar x then the gain (price minus cost) of each feeder is (x-6). With the price of Dollar 10 they can sell 20 per week.
For every dollar increase in price they will lose two sales so total sale is
(10-(x-10)) .By multiplying the number of sales of feeder with profit of each feeder sold we get total revenue y i.e.
Y=(x-6)(20-(x-10))
=(x-6) (30-x)
= -x^2+36x-180
B) compare above equation with quadratic equation we get;
a=-1 , b=36 , c= -180
A quadratic function has minimum or maximum at point -b/2a ,so revenue is increased for feeder price.
x= b/2a = 36/2(-1) = $18
AND
Maximum weekly profit is
Y(18) = (18-6) (30-18)
=$144
Answer:
the answer is A
Explanation:
Contract 2 gives the quarterback the Highest value therefore he should accept contract 2
Hey :)
I would say it’s b) Claim or argument that an error was made
Hope this helps!
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