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?

Question

Charles Anctil · Accepted Answer

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

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#

Christopher Allers · Answer

undefined To minimize cost, we need to minimize the area of the poster while still meeting the requirement for printed matter and margins.

Let the dimensions of the printed area be length (L) and width (W). The area of the printed matter is $108 cm^2$, so $L \times W = 108$.

Considering the margins, the total length and width of the poster will be:
$$L_{total} = L + 2 \times 2 = L + 4$$
$$W_{total} = W + 2 \times 6 = W + 12$$

The total area of the poster is:
$$A_{total} = L_{total} \times W_{total} = (L + 4)(W + 12)$$

The cost of the printed material is $5$ cents per $cm^2$, and the cost of the margins is $1$ cent per $cm^2$. Hence, the total cost is:
$$Cost = 5 \times 108 + 1 \times (A_{total} - 108)$$

We need to express $A_{total}$ in terms of one variable to find its minimum. Using the fact that $L \times W = 108$, we get:
$$A_{total} = (L + 4)(W + 12) = LW + 12L + 4W + 48$$
$$= 108 + 12L + 4W + 48$$
$$= 156 + 12L + 4W$$

Substituting this expression for $A_{total}$ into the cost equation gives:
$$Cost = 5 \times 108 + 1 \times (156 + 12L + 4W - 108)$$
$$= 540 + 12L + 4W - 108$$
$$= 432 + 12L + 4W$$

To minimize cost, we minimize $Cost$ with respect to $L$ and $W$. Taking partial derivatives and setting them to $0$ gives:
$$\frac{dCost}{dL} = 12 = 0 \Rightarrow L = 0$$
$$\frac{dCost}{dW} = 4 = 0 \Rightarrow W = 0$$

This is not possible because $L$ and $W$ must be positive. So, we consider the boundary values.

When $L = 108$ and $W = 1$:
$$Cost = 432 + 12(108) + 4(1) = 432 + 1296 + 4 = 1732$$

When $L = 1$ and $W = 108$:
$$Cost = 432 + 12(1) + 4(108) = 432 + 12 + 432 = 876$$

So, the least cost to make the poster is $8.76.