Question

I need help with quadratic functions, pre calculus? 

Answer 1

(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  .

Answer 2

(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 .

Answer 3

(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 .