# 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?

##### 1 Answer

The least cost occurs when we have poster dimensions of 10cm x 30cm, leading to a cost of $8.28

#### Explanation:

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 = 0 -432/(x-4)^2+12 #

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

At a min or max

# :. -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

When

# :. 6(y-12)=108 #

# :. y-12=18 #

# :. y=30 #

Hence

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

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

Hopefully you can visually confirm that a minimum does indeed occur when