Question

Find the price per unit (p) that produces the maximum profit P, if the cost function is C=4000 - 40x + .02x^2 and the demand function is p = 50-x/100

Answer 1

ok. u have
\[ \large C(x)=4000-40x+.02x^2 \]
and
\[ \large p=50-\frac{x}{100} \]
how do u define profit?

Answer 2

\[P=x \times p -c\]

Answer 3

\[P=x(500 -\frac{ x }{ 100}) - *4000 -40x + .02x ^{2}\]

Answer 4

profit is income minus cost. right?

Answer 5

yeahwhat i posted up there :)

Answer 6

sorry. i told u.

Answer 7

is it 500 or 50 in the priec equation.

Answer 8

50

Answer 9

ok so we have
\[ \large P(x)=px-C(x)=x(50-x/100)-(4000-40x+.02x^2) \]

Answer 10

yeah i got that :)

Answer 11

\[ \large P(x)=50x-x^2/100-4000+40x-.02x^2 \]
\[ \large P(x)=-.03x^2+90x-4000 \]
right?

Answer 12

yeah

Answer 13

great now find the first derivative

Answer 14

and solve 0=P'(x)

Answer 15

p=50-1500/100
p=$35

Answer 16

everything is correct.
sorry. the line crashed.

Answer 17

x=1500
p=35

Answer 18

hahah its alrite :P

Answer 19

can u help me with another one? ^^