# I need help with quadratic functions, pre calculus?

I need help with quadratic functions, pre calculus?

(a)

The number of feeder packets sold by the society is 24 per week .

Price of each feeder packet is \$10 .

The material cost for each feeder is \$6 .

For every \$1 increase in price of the feeder packet the number of sales decreases by 3 .

If x times the price \$10 increases by \$1 then x times of 3 sales will decreases .

Mathematically the  profit function

P(x) = [ Price of each feeder * number of sales per week ] - [ material cost for each feeder * number of sales per week ] .

P(x) = [ (10 + 1*x )(24 - 3*x) ] - [6(24 - 3*x)]

P(x) = [ (10 + x )( 24 - 3x) ] - [144 -18x]

P(x) = [ 240 + 24x - 30x -3x²  ] - [144 -18x]

P(x) = 240 -6x -3x² -144 +18x

P(x) = 96 +12x -3x²

So the profit function of for feeder packets is P(x) = 96 +12x -3x²  .

edited Nov 3, 2014 by bradely

(b)

The number of feeder packets sold by the society is 24 per week .

Price of each feeder packet is \$10 .

The material cost for each feeder is \$6 .

For every \$1 increase in price of the feeder packet the number of sales decreases by 3 .

If x times the price \$10 increases by \$1 then x times of 3 sales will decreases .

Mathematically the  profit function

P(x) = [ Price of each feeder packet * number of sales per week ] - [ material cost for each feeder * number of sales per week ] .

P(x) = [ (10 + 1*x )(24 - 3*x) ] - [6(24 - 3*x)]

P(x) = [ (10 + x )( 24 - 3x) ] - [144 -18x]

P(x) = [ 240 + 24x - 30x -3x²  ] - [144 -18x]

P(x) = 240 -6x -3x² -144 +18x

P(x) = 96 +12x -3x²

The profit function is P(x) = 96 +12x -3x² .

To find the maximum profit , make the first derivative of profit function to zero .

P' =  12 - 6x = 0

12 - 6x = 0

6x = 12

x = 2

So the price of each feeder packet = 10 + 1*x

= 10 + 1(2)

= 12

The price of each feeder packet is \$12 .

(c)

The number of feeder packets sold by the society is 24 per week .

Price of each feeder packet is \$10 .

The material cost for each feeder is \$6 .

For every \$1 increase in price of the feeder packet the number of sales decreases by 3 .

If x times the price \$10 increases by \$1 then x times of 3 sales will decreases .

Mathematically the  profit function

P(x) = [ Price of each feeder packet * number of sales per week ] - [ material cost for each feeder * number of sales per week ] .

P(x) = [ (10 + 1*x )(24 - 3*x) ] - [6(24 - 3*x)]

P(x) = [ (10 + x )( 24 - 3x) ] - [144 -18x]

P(x) = [ 240 + 24x - 30x -3x²  ] - [144 -18x]

P(x) = 240 -6x -3x² -144 +18x

P(x) = 96 +12x -3x²

The profit function is P(x) = 96 +12x -3x² .

To find the maximum profit , make the first derivative of profit function to zero .

P' =  12 - 6x = 0

12 - 6x = 0

6x = 12

x = 2

Now put x = 2 in  profit function , to get maximum profit .

P(2) = 96 +12(2) -3(2)²

= 96 +24 -3(4)

= 120 - 12

= 108

So the maximum profit is \$108 .