Your computer supply store sells two types of inkjet printers. the first, type a, costs $267 and you make a $24 profit on each one. the second, type b, costs $127 and you make a $20 profit on each one. you can order no more than 170 printers this month, and you need to make at least $3760 profit on them. if you must order at least one of each type of printer, how many of each type of printer should you order if you want to minimize your cost? show the entire way the problem is solved and graphed

The first, type A, costs $267 and you make a $24 profit on each one.

The second, type B, costs $127 and you make a $20 profit on each one.

number of printers = 170

Let x be the number of type A printers

Let y be the number of type B printers

x + y < = 170

you need to make at least $3760 profit on them

profit for type A is 24 and type B is 20

so 24x + 20y >= 3760

Our constraints are

x + y < = 170

24x + 20y >= 3760

x>=0  and y>=0

Cost function is C= 267 x + 127 y

Now we graph all the constraints

Now we take all the end points

corner points are

(90,80), (156.667,0) (170,0)

We plug in each point in our cost function

C= 267 x + 127 y

(90,80) => 267(90) + 127(80) = 34,190

(156.667,0) => 267(156.667) + 127(0) = 41,830

5 (170,0)=> 267(170) + 127(0) = 41,390

Here minimum cost is 34,190 at (90,80)

So we order 90 printers of type A  and 80 printers for type B in order to minimize the cost

what

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.