A printed page is to have 1 in. margin on all sides. The page should contain 80 sq. in. of type. What dimensions of the page will minimize the area of the page while still meeting the other requirements?
wh = 80
a = (w+2)(h+2) = (w+2)(80/w + 2)
= 2w + 84 + 160/w
da/dw = 2 - 160/w^2
da/dw = 0 when w^2 = 80
so, as usual a square is your best bet.
To find the dimensions of the page that will minimize the area while still meeting the other requirements, we can use optimization techniques.
Let's denote the dimensions of the page as follows:
- Width of the page: W
- Height of the page: H
We are given that the page should have a 1-inch margin on all sides. This means that the dimensions of the area containing the type (excluding the margin) are (W - 2) inches for width and (H - 2) inches for height.
The area of this type-containing region is then given by:
Area = (W - 2) * (H - 2)
We are also given that the page should contain 80 square inches of type. Therefore, we have another equation:
Area of type = 80
(W - 2) * (H - 2) = 80
Now, we need to minimize the area of the page while still meeting these requirements. To do this, we need to find the values of W and H that satisfy these equations.
To simplify the equation, we can expand the brackets:
(W - 2) * (H - 2) = WH - 2W - 2H + 4 = 80
Rearranging the equation, we have:
WH - 2W - 2H + 4 - 80 = 0
WH - 2W - 2H - 76 = 0
WH - 2(W + H) - 76 = 0
To proceed, we need to isolate one variable (either W or H). Let's isolate H:
H = (2W + 76) / W
We can substitute this expression back into the equation of the area:
Area = (W - 2) * ((2W + 76) / W - 2)
Now, we have a single-variable equation for the area in terms of W. To find the minimum area, we can take the derivative of the area with respect to W and set it equal to zero:
d(Area) / dW = 0
Solving this equation will give us the value of W that minimizes the area. Once we find W, we can substitute it back into the equation for H to find the corresponding height.