3U Functions maths question (1 Viewer)

[kaz1]

If my production costs for "n" things are given by C(x) = 2x + 1, and my income from sales is given by I(x) =12 + 7x -x^2, how many things do i sell to maximise profits?

i did I(x) > C(x) and got (-5 + sqrt77)/-2 < x < (-5 - sqrt77)/-2 and therefore you have to sell between (-5 + sqrt77)/-2 and (-5 - sqrt77)/-2 things to maximise profits.

Can any1 confirm this method/answer?

 

[tommykins]

If my production costs for "n" things are given by C(x) = 2x + 1, and my income from sales is given by I(x) =12 + 7x -x^2, how many things do i sell to maximise profits?

i did I(x) > C(x) and got (-5 + sqrt77)/-2 < x < (-5 - sqrt77)/-2 and therefore you have to sell between (-5 + sqrt77)/-2 and (-5 - sqrt77)/-2 things to maximise profits.

Can any1 confirm this method/answer?

Let P(x) = profit

P(x) = I(x) - C(x) = (12 + 7x - x²)-(2x+1) = 12 + 7x - x² -2x - 1 = -x²+5x + 11.

P'(x) = -2x + 5

Equate it to 0 to find turning points.

-2x + 5 = 0

-2x = -5
x = 5/2

When x = 5/2, profit is highest.

 

[kaz1]

Let P(x) = profit

P(x) = I(x) - C(x) = (12 + 7x - x²)-(2x+1) = 12 + 7x - x² -2x - 1 = -x²+5x + 11.

P'(x) = -2x + 5

Equate it to 0 to find turning points.

-2x + 5 = 0

-2x = -5
x = 5/2

When x = 5/2, profit is highest.

thx i was way off

 

[Senrui]

You need to answer the question properly.

You can't just sell 5/2 of a product make sure to round it up to the next whole number.