Find the area of the trapezoid.
1 answer:
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Answer:
4√10 inches wide by 6√10 inches tall
Step-by-step explanation:
"Poster area" problems have an interesting general solution. The "printed area" is maximized, or the "poster area" is minimized, when the ratio of poster dimensions is equal to the ratio of margin dimensions.
Here, the top/bottom margins total 3 inches, and the side/side margins total 2 inches. That is, the printed area will be maximized when the poster has a width : height aspect ratio of 2 : 3. If w represents the width, then the total area is ...
(w)(3/2w) = 240 . . . . square inches
w^2 = 160
w = √160 = 4√10 . . . . width in inches
Then the height is ...
height = (3/2)w = (3/2)(4√10) = 6√10 . . . . height, inches
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<em>Graphical solution</em>
A graphing calculator gives the same result for x=width. Poster width is about 12.649 inches (=4√10), and printed area is a maximum of about 170.1 square inches.
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<em>Calculus solution</em>
Let x represent the width of the poster. Then x-2 is the width of the printed area. The height of the poster will be 240/x, and the height of the printed area will be (240/x)-3. We want to maximize the printed area ...
a = (x -2)(240/x -3)
which means we want to find the value of x where the area derivative is zero.
a = 240 -480/x -3x +6
da/dx = 0 = 480/x^2 -3
Solving for x, we get ...
x^2 = 480/3 = 160
x = √160 = 4√10 . . . . . poster width, as above
