A rectangular poster is to contain 108cm^2 of printed matter with margins of 6cm at the top and bottom and 2 cm on the sides. what's the least cost to make the poster if the printed material costs 5 cents/cm^2 and the margins are 1 cent/cm^2?

Let us set up the following variables:

`{(x, "Width of poser (cm)"), (y, "Height of poster (cm)"), (P, "Printing cost (cents)") :}`

Then the dimensions of the printed matter are:

`{("Width", =x-2-2,=x-4), ("Height", =y-6-6,=y-12), ( :. " Area",=108 ,=(x-4)(y-12)) :}`

Then the dimensions (Hence Area) of the margins are:

`{(2 xx "Left/Right", =2*2*y,=4y), (2 xx "Top/Bottom", =2 * 6 * x,=12x) , (4 xx "Corners", =4 * 2*6,=48),( :. " Area", ,=4y+12x+48) :}`

Then:

`(x-4)(y-12)=108`
`:. (y-12)=108/(x-4)`
`:. y=12+108/(x-4)`

And so we can form the cost of the poster:

`C = "cost of printed at 5c" + "cost of margins at 1c"`
`:. C = 108*5 + (4y+12x+48)*1`
`:. C = 540 + 4y+12x+48`
`:. C = 588 + 4y+12x`
`:. C = 588 + 4(12+108/(x-4))+12x`
`:. C = 588 + 48+432/(x-4)+12x`
`:. C = 636+432/(x-4)+12x`

We want to minimize (hopefully) by finding `(dC)/dx`

`(dC)/dx = 0 -432/(x-4)^2+12`
`(dC)/dx = -432/(x-4)^2+12`

At a min or max `(dC)/dx=0`

`:. -432/(x-4)^2+12 = 0`
`:. 432/(x-4)^2 = 12`
`:. 36/(x-4)^2 = 1`
`:. (x-4)^2 = 36`
`:. (x-4) = +-6`
`x = 4+-6`
`:. x = -2,10`

Obviously `x>0`, so we can eliminate `x=-2`, leaving `x=10`

When `x=10 => (10-4)(y-12)=108`

`:. 6(y-12)=108`
`:. y-12=18`
`:. y=30`

Hence `x=10, y=30` (Check: `(x-4)(y-12)=6*18=108`)

With these dimensions we have:

`C = 636+432/(6)+120 = 828` cents

We should check that this value leads to a minimum (rather than a maximum) cost. As the size of the poster is finite this should really be intuitive. We could calculate the second derivative and verify that `(d^2C)/(dx)^2 > 0` when `x=10` Instead I will just use the graph `C = 636+432/(x-4)+12x`

graph{636 + 432/(x-4) + 12x [-2, 25, -2560, 2560]}

Hopefully you can visually confirm that a minimum does indeed occur when `x=10`