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# Math SL [Max/Min Problems]

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I know... I feel a bit retarded not knowing what to do, but I really don't understand... help?

Here's a question:

A rectangle is bounded by the semi-circle with equation y=(25-x^2)^(1/2), -5[u]<[/u]x[u]<[/u]5 and the x-axis. Find the dimensions of the rectangle having the largest area.

I am nowhere near the correct answer, and I don't have the right approach at all. Help?

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You need to create a differential equation and solve it. You are given y^2 + x^2 = 25 as your semicircle, meaning that it extends from the origin to a radius of 5. Now, the best way to solve this from here is to use trigonometry. You have that A = w*h, and you have that sin of the angle from the origin is (h/5), where cos of that angle is ((w/2)/5), thus h = 5sin\$, w = 10cos\$, and A = 50sin\$cos\$ = 25sin2\$.

The maximum will occur when 0 < \$ <pi, and when the derivative of 25sin\$ = 0, which is when 50cos2\$ = 0, when \$ = pi/4.

Therefore, subbing pi/4 in to the original equation, we have that the area is equal to 25, which is the square of the radius, that occurs when the rectangle has a ratio of 2w/1h and an angle from the center point to the extremities of 45 degrees. The exact dimentions will be found by using the 45 degree angle, and finding that your height will be 5 over root 2, and your width will be two times that, or 10 over root two. To check, this gives the same area of 25.

For reference, the use of derivatives can be skilled, since when we take \$ to be 45 degrees, we see that the area becomes a maximum, but you cannot prove it without derivatives. This is the most efficient method of solving the problem, though other methods exist without using trigonometry. These methods are inefficient and tend to show a general lack of understanding of derivatives, and should be avoided on tests at all costs, because it will be much harder to get part marks with a convoluted and innefficient approach if you mess up your answer. You can run in to problems interpreting the equation properly, or in solving for your differential equation, but it is by no means an issue to be unable to get this problem the first time.

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[quote name='SharkSpider' post='28760' date='Nov 21 2008, 02:18 AM']You need to create a differential equation and solve it. You are given y^2 + x^2 = 25 as your semicircle, meaning that it extends from the origin to a radius of 5. Now, the best way to solve this from here is to use trigonometry. You have that A = w*h, and you have that sin of the angle from the origin is (h/5), where cos of that angle is ((w/2)/5), thus h = 5sin\$, w = 10cos\$, and A = 50sin\$cos\$ = 25sin2\$.

The maximum will occur when 0 < \$ <pi, and when the derivative of 25sin\$ = 0, which is when 50cos2\$ = 0, when \$ = pi/4.

Therefore, subbing pi/4 in to the original equation, we have that the area is equal to 25, which is the square of the radius, that occurs when the rectangle has a ratio of 2w/1h and an angle from the center point to the extremities of 45 degrees. The exact dimentions will be found by using the 45 degree angle, and finding that your height will be 5 over root 2, and your width will be two times that, or 10 over root two. To check, this gives the same area of 25.

For reference, the use of derivatives can be skilled, since when we take \$ to be 45 degrees, we see that the area becomes a maximum, but you cannot prove it without derivatives. This is the most efficient method of solving the problem, though other methods exist without using trigonometry. These methods are inefficient and tend to show a general lack of understanding of derivatives, and should be avoided on tests at all costs, because it will be much harder to get part marks with a convoluted and innefficient approach if you mess up your answer. You can run in to problems interpreting the equation properly, or in solving for your differential equation, but it is by no means an issue to be unable to get this problem the first time.[/quote]

mmm i gotta say it's pretty good explanation, though an IB math SL student might get a little confused...especially since you used that horryfying word "differential equations". Dont' do that! No but seriously if you understand derivitaves using this type of implicit diff is the best way of solving the problem. Most people probably learned to solve this by doing an optimization problem, which is what this is. Generally, it occurs where your derivate equals zero...though most people can't see at first glance the derivate of what cause sometimes its hard to find the equation. In this case the easy way out would be to take the derivate of Y which is a pain in the ass and takes forever compared to just using implicit (or even further just take the negative partial of F in terms of x over the partial in terms of y, Multivariable Calculus material).

you need to identify your parameters to know what to take the derivative for. Sharkspider gave an awsome, probably the best, two parameters but he went through polar coordinates and then back to get the answer. What you could've done is set the area to equal X (where X is half the base of our final rectangle; half because we can only maximize one side of he y-axis at a time) times height where height equals F(x) which equals sqrt(25-x^2). Take the derivate (pain in the ass as you see) and get sqrt(25-x^2) + x(-x/sqrt(25-x^2)) simplifyies to (25 - 2x^2)/sqrt(25-x^2).

So rate of change of the area = (25 - 2x^2)/sqrt(25-x^2). You want the max, so set the deriv. equal to zero and obvserve a change in sign. X would equal 5/sqrt(2) just like SharkSpider said, meaning that your BASE equals that times two (remember that we maximized only half of the rectangle). You still need a height, but we defined height as F(x) which is a fancy name fo Y. So F(5/sqrt 2) = 5/sqrt 2 which makes sense*. So base equals 10/sqrt 2 times height equals 5/ sqrt 2 equals exactly 25 square units.

*what I meant by it makes sense is that generally speaking, maximizing area wil ALWAYS yield a perfect square area, ie, 25. You could've predicted that the area was gonna be 25 as soon as you saw the radius of the semi-circle to be 5, but these tricks come with practce, no need to know them.

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Oh... I see... thanks! *I got the question already, but yeah, I just remembered I asked the question ages ago*

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Now, calculus is a fine way to approach this problem, don't get me wrong, but there is an extremely simple solution.

Let (a,b) be he coordinate of the point on the rectangle in quadrant 1 touching the semicircle.

We know a^2+b^2=25, and you want to maximize the area, which is equal to 2ab.

(a-b)^2 >= 0

a^2+b^2 >=2ab

25 >=2ab=Area of rectangle

Equality holds when (a-b)^2=0, or when a=b. Since a^2+b^2=25, we have a^2=25/2 or a=b=5/sqrt(2).

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[quote name='rofler' post='30107' date='Dec 11 2008, 12:53 AM']Now, calculus is a fine way to approach this problem, don't get me wrong, but there is an extremely simple solution.

Let (a,b) be he coordinate of the point on the rectangle in quadrant 1 touching the semicircle.

We know a^2+b^2=25, and you want to maximize the area, which is equal to 2ab.

(a-b)^2 >= 0

a^2+b^2 >=2ab

25 >=2ab=Area of rectangle

Equality holds when (a-b)^2=0, or when a=b. Since a^2+b^2=25, we have a^2=25/2 or a=b=5/sqrt(2).[/quote]

mmm, not bad except it's a little hard to follow your reasoning but I get what you're saying. The only problem is that the only way this would work is if you knew that a has to equal b in order to have maximum area. For math SL optimization problems it's much better just to take the derivative and set it equal t zero. But nice proof neways.

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It's not really a proof, it's simply taking the fact that we already proved the greatest cosine/sine sum occurs at fourty five degrees and using it to solve the equation. On a test, that wouldn't be considered a proof because you get a = b without doing any work.

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Proof: I claim that the maximum of the area is 25. To do this, we will show first that the area is <= 25, and then show when equality holds.

Let the upper right most vertex of the rectangle have form (a,b)

Area=2ab=a^2+b^2-(a-b)^2<=a^2+b^2=25

Equality holds when (a-b)^2=0 which is equivalent to a=b. Since a^2+b^2=25, 2a^2=25 which gives a=5/sqrt(2), so a=b=5/sqrt(2) has to be the vertex of the upper right most vertex for equality to hold.

Touche P.S. just because it's a simpler proof doesn't make it invalid... Edited by rofler

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[quote name='rofler' post='30136' date='Dec 11 2008, 11:23 PM']Proof: I claim that the maximum of the area is 25. To do this, we will show first that the area is <= 25, and then show when equality holds.

Let the upper right most vertex of the rectangle have form (a,b)

Area=2ab=a^2+b^2-(a-b)^2<=a^2+b^2=25

Equality holds when (a-b)^2=0 which is equivalent to a=b. Since a^2+b^2=25, 2a^2=25 which gives a=5/sqrt(2), so a=b=5/sqrt(2) has to be the vertex of the upper right most vertex for equality to hold.

Touche P.S. just because it's a simpler proof doesn't make it invalid...[/quote]

Ya, in fact it makes it better, but it's still simpler to use calculus. Maybe not faster, but definately simpler.

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