Comments on: Show that the area of the triangle enclosed by the segment is largest when a = b? http://www.thearealist.com/triangle-area/show-that-the-area-of-the-triangle-enclosed-by-the-segment-is-largest-when-a-b Fri, 03 Feb 2012 13:07:59 -0500 http://wordpress.org/?v=2.8.4 hourly 1 By: Duke http://www.thearealist.com/triangle-area/show-that-the-area-of-the-triangle-enclosed-by-the-segment-is-largest-when-a-b/comment-page-1#comment-1708 Duke Fri, 22 Jan 2010 10:13:59 +0000 http://www.thearealist.com/triangle-area/show-that-the-area-of-the-triangle-enclosed-by-the-segment-is-largest-when-a-b#comment-1708 The Area of the triangle is A = (1/2)ab We know that a and b are related by a^2 + b^2 = 20^2 Use this to eliminate a from the area equation since a^2 = 400-b^2 or a = sqrt(400-b^2) So A = (1/2)* [sqrt(400-b^2)]*b Now we want the Area A to be a max. So take the derivative dA/db and set equal to zero: A '(b) = (1/2)*{ sqrt(400-b^2) + b*(1/2)(-2b)/sqrt(400-b^2) } using the product and chain rule. Set this = 0 and solve for b: you get b = 20/sqrt(2) then a = sqrt(400-b^2) = sqrt(400/2) = 20/sqrt(2) Therefore, a = b for the maximum area. (I'm assuming you're in first year single variable calculus; in multivariable calculus, this can be done even quicker by other methods)<br><b>References : </b><br> The Area of the triangle is A = (1/2)ab

We know that a and b are related by a^2 + b^2 = 20^2
Use this to eliminate a from the area equation since
a^2 = 400-b^2 or a = sqrt(400-b^2)

So A = (1/2)* [sqrt(400-b^2)]*b
Now we want the Area A to be a max. So take the derivative dA/db and set equal to zero:
A ‘(b) = (1/2)*{ sqrt(400-b^2) + b*(1/2)(-2b)/sqrt(400-b^2) }
using the product and chain rule.
Set this = 0 and solve for b: you get
b = 20/sqrt(2)
then a = sqrt(400-b^2) = sqrt(400/2) = 20/sqrt(2)

Therefore, a = b for the maximum area.

(I’m assuming you’re in first year single variable calculus; in multivariable calculus, this can be done even quicker by other methods)
References :

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