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    confused's Avatar
    confused Posts: 1, Reputation: 1
    New Member

    Mar 18, 2004, 03:14 PM
    Right Triangle
    How do you solve this using the following formulas?

    A right triangle has a hypotenuse of 12 cm. Find the lengths of the legs A and B so that the area of the triangle will be a maximum. Find the area. Round the answers to the nearest hundredth.

    A =1/2 AB and A^2 + B^2 = 12^2
    eawoodall's Avatar
    eawoodall Posts: 230, Reputation: 5
    Full Member

    Mar 26, 2005, 12:46 AM
    max area of right triangle
    12 ^ 2 = 144.
    A^2 + B^2 = 144.
    1/2 AB = area
    you want max area.
    plug in numbers.
    let A = 10, B = (44)^(1/2)
    1/2 * 10 * 6.6332 = 33.16

    let A=B,
    144/2 = 72, so A = 72^(1/2) = B = 8.49
    1/2 * A * B = 1/2 * 72 = 36.

    area is max when length of shorter two sides is equal, i.e. 90/45/45 triangle.
    Dr_Calculus's Avatar
    Dr_Calculus Posts: 35, Reputation: 1
    Junior Member

    Mar 27, 2005, 02:17 PM
    easier way
    An easier way to solve this problem is to use a bit of intuition. A square is a rectangle with the greatest amount of area for fixed amounts of material (for example if you had x amount of fence and wanted the greatest amount of area you could enclose in a meadow), and since triangles are directly proportionate to squares/rectangles, the greatest area triangle will be 1/2 of the square with diagonal length of 12. Thus, we know that it is a 45-45-90 triangle and thus know that A=B=12/sqrt(2)
    reinsuranc's Avatar
    reinsuranc Posts: 92, Reputation: 6
    Junior Member

    Apr 11, 2005, 02:04 PM
    Without calculus
    Quote Originally Posted by Dr_Calculus
    A square is a rectangle with the greatest amount of area for fixed amounts of material.
    We don't know what subject the person who posed this problem is taking.

    The above quote is of course true, but it results from a calculus proof; perhaps the person who posed the problem is taking a pre-calculus course.

    The first person who replied seems to have used trial and error. There's nothing wrong with that, but it is a little unsatisfying.

    Is there a solution that is more logical than trial and error, but that does not require a result from calculus?
    eawoodall's Avatar
    eawoodall Posts: 230, Reputation: 5
    Full Member

    Apr 11, 2005, 06:00 PM
    no matter what method you use to determine what values you want to use for your length of sides A and B, you have to eventually test those results to see if you got what you wanted. Testing the extrema (limits) allows one easy method for knowing if you got the answer you seek.

    you may construct a graph
    where the X axis domain is (0,12]
    where the y axis range is (0,12]
    and the graph plots a equation where points X^2 +Y^2 = 144
    because traingles have positive lengths of sides we exclude any part of the graph X<0 or Y<0. i.e. we only include X>0 and Y>0 because triangles has strictly positive lengths of all three sides.
    plot the graph.

    12 ^ 2 = 144.
    A^2 + B^2 = 144.
    1/2 AB = area
    you want max area.
    plug in numbers of limits or a number close to the limit that is easy to do.
    limit as A goes to zero implies B approaches C.
    so we let A = 1 to test area near the limit of A appraoching zero.

    let A = 1, B = (143)^(1/2)
    1/2 * 1 * 11.958 = 5.98
    we notice small area.

    let A = (143)^(1/2), B = 1
    1/2 * 11.958 * 1 = 5.98
    notice same area as first result.
    so when A = 1 or B = 1 we get same area.
    seems reflexsive of length of sides A and B with respect to area.

    let A = 2, B = (140)^(1/2)
    1/2 * 2 * 11.832 = 11.832
    notice area got larger as length of A increased in value with respect to B.

    et cetera...

    finally we see when A=B we get results we seek. See earlier post.

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