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Student in a college prepare a poster for display . The poster is to contain 200 sq inches of printed matter , with margins of 4 inches at top and bottom and of 2 inches on each side. Find the dimensions of the poster if the total area is to be minimum.
Okay, this is an optimization problem.
The first thing you are going to want to do is draw a picture - it really helps you visualize it. I set the length equal to y and the width equal to x. From the picture, you should be able to see that the length is equal to y-8 (eight being the size of the margins). You should also realize that the width is equal to x-4.
Put this into the equation for area, A=x*y and you should get A=(y-8)(x-4)
This doesn't help yet, but remember that A=x*y, and we want the printed area to be 200 sq inches. So, plug 200 in for A, and you get 200=x*y. Solve this for y, and you get y=200/x.
Now plug that into your first area equation, and you get A=((200/x)-8)(x-4).
From there, you should be able to simplify it and take the derivative. Once you have the derivative, it's simply a matter of optimizing the problem. Look for the critical points on the graph, and find the local maximum. Once you have a value for the local max, you should be able to return to the equation y=200/x. Just plug in the value of the relative maximum you found for x, and solve for y. From there, you should get your dimensions!
Good luck!