Max Possible Area

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  1. Tycho's Avatar
    1 10-07-2012  16:42
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    Max Possible Area
    Q: Find the maximum possible area of a right-angled triangle which has hypotenuse of length 5m.

    Is this an optimisation question? I can get close to the answer by trial-and-error, but how do you solve this mathematically?
  2. raheem94's Avatar
    2 10-07-2012  16:57
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    Re: Max Possible Area
    (Original post by Tycho)
    Q: Find the maximum possible area of a right-angled triangle which has hypotenuse of length 5m.

    Is this an optimisation question? I can get close to the answer by trial-and-error, but how do you solve this mathematically?
    I get the answer as 6.25, is it the answer?
  3. Groat's Avatar
    3 10-07-2012  17:01
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    Re: Max Possible Area
    (Original post by Tycho)
    x
    Label the two unknown sides x and y. We know that x^2 + y^2 = 25 as it's a right-angled triangle. We can arrange that to get y = \sqrt{25 - x^2}

    The area of a triangle is \frac{1}{2}bh so we plug our values in: A = \frac{1}{2}x\sqrt{25 - x^2}

    Then you differentiate with respect to x.
    Last edited by Groat; 10-07-2012 at 17:04.
  4. theandyguthrie's Avatar
    4 10-07-2012  17:03
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    Re: Max Possible Area
    Use x and y as the side values

    express the area in terms of only x or y (using x^2 + Y^2 = 25)

    differentiate and find the value of x at which a maximum occurs.

    Solve for Y.

    Sub back into original area equation.


    I can show you my working if you would like - but people don't like complete solutions.
  5. raheem94's Avatar
    5 10-07-2012  17:04
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    Re: Max Possible Area
    (Original post by Groat)
    Label the two unknown sides x and y. We know that x^2 + y^2 = 25 as it's a right-angled triangle. We can arrange that to get y = \sqrt{25 - x^2}

    The area of a triangle is \frac{1}{2}bh so we plug our values in: A = \frac{1}{2}x\sqrt{25 - x^2}

    Then you differentiate with respects to x.
    I did exactly the same thing to get the answer but the problem is that the question is labelled 'secondary school' while this method requires C3 differentiation.
  6. Groat's Avatar
    6 10-07-2012  17:08
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    Re: Max Possible Area
    Alternative method for people who have done more A-Level maths.

    You can think of the triangle in terms of \theta and the hypotenuse (5).

    The base is equivalent to 5cos(\theta) and the height is equivalent to 5sin(\theta).

    The area is then: A = \frac{25}{2}cos(\theta) \ sin(\theta) and you differentiate with respect to \theta .
  7. SubAtomic's Avatar
    7 10-07-2012  17:10
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    Re: Max Possible Area
    (Original post by Tycho)
    Q: Find the maximum possible area of a right-angled triangle which has hypotenuse of length 5m.

    Is this an optimisation question? I can get close to the answer by trial-and-error, but how do you solve this mathematically?
    Hypotenuse is \sqrt {a^2+b^2}

    So hyp is 5 then h^2=25

    So a^2+b^2=25

    a=3, b=4.

    Area of a triangle \frac{1}{2}ab

    Don't think it is optimisation because as far as I am aware a right triangle with a predetermined hypotenuse will have the same area no matter how much you faff.
    Last edited by SubAtomic; 10-07-2012 at 17:14.
  8. Groat's Avatar
    8 10-07-2012  17:11
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    Re: Max Possible Area
    (Original post by raheem94)
    I did exactly the same thing to get the answer but the problem is that the question is labelled 'secondary school' while this method requires C3 differentiation.
    Maybe they just want you to spot that an angle of 45 degrees will maximise the area?
  9. raheem94's Avatar
    9 10-07-2012  17:19
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    Re: Max Possible Area
    (Original post by Groat)
    Maybe they just want you to spot that an angle of 45 degrees will maximise the area?
    I haven't done GCSE so don't know much about 'secondary school' kids knowledge. But i don't feel a GCSE candidate will be able to spot it.
  10. raheem94's Avatar
    10 10-07-2012  17:20
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    Re: Max Possible Area
    (Original post by SubAtomic)
    Hypotenuse is \sqrt {a^2+b^2}

    So hyp is 5 then h^2=25

    So a^2+b^2=25

    a=3, b=4.

    Area of a triangle \frac{1}{2}ab

    Don't think it is optimisation because as far as I am aware a right triangle with a predetermined hypotenuse will have the same area no matter how much you faff.
    Your solution gives area as 6, while my answer is 6.25.

    You certainly can't get the max area in this way, read the other posts.
  11. Groat's Avatar
    11 10-07-2012  17:24
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    Re: Max Possible Area
    (Original post by raheem94)
    I haven't done GCSE so don't know much about 'secondary school' kids knowledge. But i don't feel a GCSE candidate will be able to spot it.
    Think about triangles in a semi-circle with diameter 5. The biggest area is when the triangle is isosceles (or when \theta is equal to 45 degrees).
  12. raheem94's Avatar
    12 10-07-2012  17:29
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    Re: Max Possible Area
    (Original post by Groat)
    Think about triangles in a semi-circle with diameter 5. The biggest area is when the triangle is isosceles (or when \theta is equal to 45 degrees).
    Yeah good spot.

    Though i would have done it in the following way if differentiation wasn't allowed, \displaystyle \frac{25}2 \cos \theta \sin \theta = \frac{12.5}2 \times 2 \sin \theta \cos \theta = \frac{12.5} 2 \times \sin ( 2 \theta )

    \text{Max } \sin ( 2 \theta ) = 1 \implies \theta = 45^{\circ}
    Last edited by raheem94; 10-07-2012 at 17:31.
  13. gearoid94's Avatar
    13 10-07-2012  17:30
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    Re: Max Possible Area
    I suppose the easiest way to do it would be to take the hypotenuse as the base. Then the greatest area will be when the height is the greatest, ie when the 2 other angles are 45 degrees. Sounds more like GCSE reasoning.
  14. Hopple's Avatar
    14 10-07-2012  17:32
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    Re: Max Possible Area
    (Original post by Tycho)
    Q: Find the maximum possible area of a right-angled triangle which has hypotenuse of length 5m.

    Is this an optimisation question? I can get close to the answer by trial-and-error, but how do you solve this mathematically?
    This is a bit of a strech for secondary school rather than A Level, but it doesn't use differentiation at least. As above, the area is 0.5*x*sqrt(25-x^2) where x is the length of one of the sides (or both ). Square it and ignore the 0.25 to get (25-x^2)*x^2. I think GCSE can find maximum values of quadratics by writing them as (z+a)^2+b, and this quartic is essentially a quadratic in x^2. However, this is still a stretch for GCSE.

    Edit: the semi-circle thing seems much more reasonable.
    Last edited by Hopple; 10-07-2012 at 17:33.
  15. Hopple's Avatar
    15 10-07-2012  17:35
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    Re: Max Possible Area
    (Original post by raheem94)
    Yeah good spot.

    Though i would have done it in the following way if differentiation wasn't allowed, \displaystyle \frac{25}2 \cos \theta \sin \theta = \frac{12.5}2 \times 2 \sin \theta \cos \theta = \frac{12.5} 2 \times \sin ( 2 \theta )

    \text{Max } \sin ( 2 \theta ) = 1 \implies \theta = 45^{\circ}
    I don't think we learnt compound angles until A Level, even then it was C3 (I think).
  16. raheem94's Avatar
    16 10-07-2012  17:38
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    Re: Max Possible Area
    (Original post by Hopple)
    I don't think we learnt compound angles until A Level, even then it was C3 (I think).
    Yes, it is C3, was just suggesting an alternate technique.
  17. SubAtomic's Avatar
    17 10-07-2012  17:44
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    Re: Max Possible Area
    (Original post by raheem94)
    Your solution gives area as 6, while my answer is 6.25.

    You certainly can't get the max area in this way, read the other posts.
    If it is A-level, not even sure if optimisation gets taught in the higher GCSE syllabus. My solution is basically a GCSE answer as far as I am aware. Square numbers and all. If you can optimise the area with pre sixth form knowledge then fair enough.

    Suppose it is like groat said, you'd have to notice that the isoc triangle gives the max area and go from there. Which if in context will be second nature depending on what's been going on in class.
    Last edited by SubAtomic; 10-07-2012 at 17:55.
  18. aznkid66's Avatar
    18 11-07-2012  05:02
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    Re: Max Possible Area
    Surely they are supposed to know about how squares maximize areas?

    Let's say that a rectangle with the smaller side x and the larger side x+k. The perimeter would be P=2(x)+2(x+k)=4x+2k. Thus, x (in terms of P and k) would be would be x=(P-2k)/4, and x+=(P+2k)/4

    Now, the area A=(P-2k)/4 * (P+2k)/4=(P^2-4k^2)/16

    Now we want to maximize Area by changing k but not P. So let's look at what happens when we increase/decrease k.

    As k increases from 0, k^2 also increases. When k^2 increases, P^2-4k^2 decreases (because you're subtracting more) and thus as k increases, the Area decreases.

    As k decreases from 0, k^2 still increases, because (-k)^2 = k^2. Thus, as k decreases, the Area decreases.

    So, since we want to subtract the least from P^2, what value of k do we want? That value is when we subtract nothing, aka when 4k^2=0, aka when k=0. Therefore, we don't want any difference between the two sides; we want them to be equal.

    The same reasoning works for a(n isosceles) triangle, except that the area is halved.
  19. Tycho's Avatar
    19 11-07-2012  09:50
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    Re: Max Possible Area
    Thanks all for your help. I managed to get the same answer as you, just by trial and error rather than the ways you have. I tried some of the ideas you mentioned but wasn't able to follow through as accurately as you.

    The reason I marked as secondary school is because in Scotland we do A-level (Highers & Advanced Highers) qualifications in secondary school. Apologies for any confusion.
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