Probem31+Deborah+Sylvia

For this problem, what we are trying to do is minimize the area of the poster. It is given that the top/bottom margins and side margins are 6 cm and 4 cm, respectively, and that the area of printed material is fixed at 384 cm^2. Basically, what we are dealing with is a rectangle of printed material with 4 inches of blank space on either side of it and 6 inches of blank space above and below it, creating a larger rectangle which the printed rectangle is inside of. If we say that the dimensions of the outside rectangle's dimensions are L by W, then the formula for the area will be LW (according to bh = A). Then, because we know the area of the smaller triangle must be equal to 384, we can determine that the dimensions of the smaller rectangle are (L - 12) by (W - 8), taking into account the 6 inches on the top and bottom and 4 inches on either side. We have to isolate one variable in order to solve, so solving for L we will get that L = [384/(W - 8)] + 12. We then plug this equation in as L in the first area equation, giving us A = W[(384/W-8) + 12], or A = 384W/(W - 8) + 12W, after distributing. To find the minimum, we have to take the derivative and set it equal to 0 (finding the horizontal tangent line). This gives us A' = [(W - 8)(384) - 384W]/(W - 8)^2 + 12 = 0 After distribution, the 384Ws on top cancel out, the 12 shifts to the other side as a negative, and after some calculation we find that W = 24. ^^ Minimum = 8 (taking into account both 4 inch margins) Maximum = ?

You need to consider what the absolute smallest value we have for W and what is the largest value. Consider that W includes the 4" on either side of the printed area, so the minimum cannot equal zero! As for the largest value, you might want to examine what the absolute smallest value for L is, and how W reacts at that value of L! In this case, of course, W (and L) cannot actually equal these smallest values (you would have NO printed area) so your domain will have curvy endpoints, so you will need to examine the limit at each of these values.

My (Sylvia's) work: I did something different .

Ainside = (L)(W) = 384 L = 384/W

Atotal = (total L) ( total W) Atotal = (L + 12) (W + 8)

substitute L with 384/W Atotal = (384/W + 12) (W + 8)

simplify Atotal = 384 + 3072/W + 12W + 96 Atotal = 12W + 3072(W^-1) + 480

Take derivative dA/dW = 12 - 3072/(W^2)

Critical points! 12 - 3072/W^2 = 0 12 = 3072/W^2 W^2 = 3072 / 12 W = +/- 16 (but we know it can't be negative because we don't have negative centimeters of posters)



So, when W = 16, Atotal is at a minimum

Atotal = 12W + 480 + 3072/W Atotal = 12(16) + 480 + 3072/16 = 864

Total minimum Area = 864 cm^2