Problem: Optimization A printed page is to contain 50 square inches of printed material. There are to
be 4-inch margins at the top and bottom, and 2-inch margins on each side. What are the dimensions of
the page will give a page with the minimum area?
We are to minimize the area of the page. Thus we need a function which gives the area of the page.
Let W be the width of the page in inches. Let L be the length of the page in inches. Let A = LW be the
area of the page in square inches. This gives the area of the page as a function of two variables.
We need to write Area A, as a function of 1 variable. We use the fact that the area of the printed
material is 50 inches to put L in terms of W. The width of the printed material is W - 4 inches. The length
of the printed area is L 8. This gives us (W - 4)(L - 8) = 50 and so (L - 8)
=
So now, we can write A as a function of one variable. So A = LW, becomes A =
The Domain of A is : 4 < W <
or , L =
+ 8.
+ W 8W.
since must be positive . Next we find the critical numbers of A.
A'(W) =
=
+8
+8 =
+8
A(W) is not defined at x = 4, but 4 is not in the domain of A.
L-8
W-4
L
A(W) =
8
Thus we dont use x = 4 as a critical number.
+ 8 = 0 when 8 =
or when
= 200. This gives 8(W2 - 8W + 16) = 200.
From this we get 8W2 - 64W + 128 = 200
or 8W2 - 64W - 72 = 0.
W
Factoring, we get 8(W 9)(W + 1) = 0, which gives W = 9, and W = -1.
However, -1 is not in the domain of A, so we use only W = 9.
We substitute W = 9 in (W 4)(L - 8) = 50, to find L. We get (9 4)(L 8) = 50,
or 5(L 8) = 50 or L - 8 = 10 or L = 18.
We use the first derivative test to verify W = 9 gives a minimum area. Using test point W = 5 to the left
of 9 and test point W = 10 to the right of 9, we get A(W) < 0 on (4, 9), A(W) > 0 on (9, ), so by the
first derivative test, the area of the page is minimized if W = 9.
ANS: The dimensions of page are: W = 9 inches wide and L =18 inches long.