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]}