Like

Report

A poster is to have an area of $ 180 in^2 $ with an $ 1 $-inch margins at the bottom and sides and a $ 2 $-inch margin at the top. What dimensions will give the largest printer area?

$2 \sqrt{30}$ in $\times \frac{90}{\sqrt{30}}$ in

You must be signed in to discuss.

Oregon State University

Harvey Mudd College

Baylor University

Idaho State University

reason. Yes, no, we're told. A poster is to have an area of 100 in square inches with one inch margins of the bottom Insides and a two inch margin at the top were asked what dimensions will give the largest printer area. They sent me home, so I'll draw a quick well figure. Mhm. Understand yourself. So we have a margin of two here. Margins of one here, no length. This side is X minus two, and his height is why minus three. Yeah, the total height we're saying is why in the total length is X My mistake. I should have said that first, but it follows that area of the entire poster, which is 180 square inches. This is X Times y. So we have that y is equal to 1 80 over X, and the area of the smaller printed part shall call a is X minus two times Y minus three. We can write this as a function of X A N X equals X minus two times 180 over X minus three. What is this? Some? Yeah, I'll distribute. This is equal to 186 minus three X Never minus 360 over X. Now, in order to maximize the printer area, I'm going to take the derivative. Mm. We'll find the absolute max of a now notice that a of X approaches, right? We're not negative. Infinity as X approaches zero from the right and a f X approaches Negative infinity. As X approaches infinity, it follows that the local maximum on zero infinity is also an absolute Yeah, maximum on this interval. Yeah. Now let's find the critical numbers. So if we differentiate a prime of X supposed to be is equal to 20 Let's see Negative three plus 360 over X squared. We set this equal to zero. So we have X squared equals 3 60/3, which is 120. So that X is equal to plus or minus two or 30. However, were only look interested in positive X so X is equal to positive 2. 30. So the absolute max occurs at 2. 30 first grade. So what? Yes, Shoot Mhm, he said Okay. You can use this to find why. So why is equal to 1 80 over x this is 1 80/2 30 she was 90/30 which is the same as three times Route 30 and therefore, the dimensions of the poster with largest printed area are starting with X to Route 30 and this is in inches by three. Route 30 also in inches, which this is approximately 10.95 inches by 16.43 inches, it's