Calculus I Optimization

Question

The top and bottom margins of a poster are a  cm and the side margins are each b  cm. If the area of printed material on the poster is fixed at c  square centimeters, find the dimensions of the poster with the smallest area.

Answer 1

Let the dimensions of the material be w (width) and h (height).  Since there is a margin of a above and below the material, the height of the poster is H = h + 2a.  Likewise, the width of the poster is W = w + 2b.

 

Since the area of the printed material is c, we have wh = c.

 

The area of the poster is A = WH.  Substitution yields

A = (w + 2b)(h + 2a). 

 

From wh = c, we have w = c/h.

 

A = (c/h + 2b)(h + 2a)

= c + 2ac/h + 2bh + 4ab.

 

Keep in mind that h is the variable, and the other letters are constants.  We now find critical points by differentiation:

 

A' = -2ac/h² + 2b.  Setting the derivative to 0,

0 = -2ac/h² + 2b, or

h = √(ac/b).  The negative root is discarded, since h≥0.

 

We use the second derivative test to see if this is a min or max:

A'' = 4ac/h³.  Since h>0, A''>0 and we have a local min for A.  The interval for h is [0,∞).  A approaches ∞ as h→0⁺, due to the 2ac/h term.  Likewise, A→∞ as h→∞ due to the 2bh term.  The local min is the absolute min.

 

So h=√(ac/b) minimizes A.  We also have w = c/h, or w=c√(b/ac) = √(bc/a).

 

The minimum area is then A = WH = (w + 2b)(h + 2a).  Substituting in our solutions for w and h yields

A = [√(bc/a) + 2b] [√(ac/b) + 2a].