# Question #5cf1a

##### 1 Answer

The smallest area is

#### Explanation:

Let us set up the following variables:

# {(x, "Width of poster (cm)"), (y, "Height of poster (cm)"), (A, "Area of poster ("cm^3")") :} #

Then the dimensions of the **printed matter** are:

# {("Width", =x-6-6,=x-12), ("Height", =y-4-4,=y-8), ( :. " Area",=384 ,=(x-12)(y-8)) :} #

So we have;

# 384 =(x-12)(y-8) #

# :. y-8 = 384/(x-12)#

# :. y = 8+384/(x-12)#

# " " = (8(x-12)+384)/(x-12)#

# " " = (8x-96+384)/(x-12)#

# " " = (8x+288)/(x-12) \ \ \ \ \ ..... (star)#

And the total area of the poster is given by:

# A = xy #

# \ \ \ = (x)((8x+288)/(x-12)) \ \ \ # (using (#star# ))

# \ \ \ = (8x^2+288x)/(x-12) #

We want to minimize (hopefully) by finding

# (dA)/dx = { (x-12)(16x+288) - (1)(8x^2+288x) } / (x-12)^2 #

# " " = { 16x^2+288x-192x+3456-8x^2-288x } / (x-12)^2 #

# " " = { 8x^2-192x+3456 } / (x-12)^2 #

At a min or max

# :. { 8x^2-192x+3456 } / (x-12)^2 = 0 #

# :. 8x^2-192x-3456 = 0 #

# :. 8(x^2-24x-432)=0 #

# :. x^2-24x-432 = 0 #

# :. (x+12)(x-36) = 0 #

This equation leads to the two solutions:

# x= -12# , or# x=36#

Obviously

# y = (288+288)/(36-12) \ \ \ \ \ # (using (#star# ))

# \ \ = 576/24 #

# \ \ = 24 #

With these dimensions we have:

# A = 36*24 #

# \ \ \ = 864 #

We should check that this value leads to a minimum (rather than a maximum) area, which we confirm graphically:

graph{(8x^2+288x)/(x-12) [-2, 50, -2000, 2000]}