Area List

If you have a right-angled triangle with area = S, which one would have the smallest hypotenuse?

Let x and y be the legs of the right triangle. Then the hypotenuse z satisfies

z² = x² + y².

Since z>0, minimizing z² will minimize z.

The area of the triangle is

S = (1/2)xy
→
2S/x = y
→
z² = x² + (2S/x)².

So find a minimum for the function

f(x) = x² + (2S/x)²:

f’(x) = 2x + 4S^2 (-2 x^(-3))
= 2x – 8S² x^(-3).

This does not exist when x=0, but if x=0 we don’t have a triangle. Set to 0:

0 = 2x – 8S² x^(-3)
Multiply through by x^3:
0 = 2x^4 – 8S²
0 = x^4 – 4S²
4S² = x^4
[4S²]^(1/4) = x
√(2S) = x.

(You need to perform the second derivative test to show x = √(2S) gives a minimum.)

Then y = 2S/x = 2S/√(2S) = √(2S),

and

z² = (√(2S))² + (√(2S))²
= 2S + 2S
= 4S

–>

z = 2√S is the smallest hypotenuse.

Hope this helps.

♣ ♣

One Response

1. ♣ K-Dub ♣ Says:
   November 11th, 2009 at 7:23 pm

   Let x and y be the legs of the right triangle. Then the hypotenuse z satisfies

   z² = x² + y².

   Since z>0, minimizing z² will minimize z.

   The area of the triangle is

   S = (1/2)xy
   →
   2S/x = y
   →
   z² = x² + (2S/x)².

   So find a minimum for the function

   f(x) = x² + (2S/x)²:

   f’(x) = 2x + 4S^2 (-2 x^(-3))
   = 2x – 8S² x^(-3).

   This does not exist when x=0, but if x=0 we don’t have a triangle. Set to 0:

   0 = 2x – 8S² x^(-3)
   Multiply through by x^3:
   0 = 2x^4 – 8S²
   0 = x^4 – 4S²
   4S² = x^4
   [4S²]^(1/4) = x
   √(2S) = x.

   (You need to perform the second derivative test to show x = √(2S) gives a minimum.)

   Then y = 2S/x = 2S/√(2S) = √(2S),

   and

   z² = (√(2S))² + (√(2S))²
   = 2S + 2S
   = 4S

   –>

   z = 2√S is the smallest hypotenuse.

   Hope this helps.

   ♣ ♣
   References :

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