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In the following exercises, solve the given maximum and minimum problems.A rectangular hole is to be cut in a wall for a vent. If the perimeter of the hole is 48 in. and the length of the diagonal is a minimum, what are the dimensions of the hole?

Calculus 1 / AB

Chapter 24

Applications of the Derivative

Section 7

Applied Maximum and Minimum Problems

Derivatives

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question 13 Uh, here were given. That area equals 49 square. When we have area, we know that our to Gary Bulls are both positive. So we know that lane finds or it's weird way also are know that our perimeter is our maximum and the formula for perimeter of a rectangle is to l plus two w I'm gonna solve my perimeter, sir. I'm going to solve my area. My length, times with himself for l. I have l u 49 over w. And I'm just ignoring the feet square right now. Um, and I know my answer will be in feet. So now I'm gonna take this equation here more racy and l sub secure in my perimeter. I'm gonna have a perimeter two times. Wants to get you. So this gives me a perimeter. Walls 98 we'll double you plus double. And now, um, I want to, but I know that perimeter is my maximum so that when I take the derivative, I'm gonna get a zero. But first I like to write the w. I'd like to write the denominator as a negative exponents. I'm just gonna write groom ITER equals 98 of new negative plus two. And now when I take my derivative, I am get Oh zero equals because my maximum negative 98 that's I have a 98 over W Square plus two. And when I saw or my w attractive moods I thio weird thank you to use squared two and you were 49 thing 49 plus reminding us in this case we're only looking at areas has value. So looking at seven and my units are in the my with his seven, my length would also have to eat seven because seven times that, you know

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