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Sarah_Marie · Oct 22, 2010, 01:58 PM

A triangle's base and height have a sum of 36 cm. Find the base and height of the triangle such that its area is a maximum possible area.

harum · Oct 22, 2010, 02:15 PM

If the height is h and the base is b, then irrespective of the exact geometry the area of the triangle is: S = (1/2)*b*h.
If a sum of the height and the base is fixed, b+h=L, then h = L - b, and S(b) = (1/2)*b*(L-b) = (1/2)*b*L - (1/2)*b^2.
To find the value of b for which S reaches maximum: dS/db = (1/2)*L - b = 0.
b = L/2;
To make sure that b = L/2 is the maximum, double check that the second derivative is negative: d^2S/(db)^2 = -1.

galactus · Oct 23, 2010, 05:42 AM

Just as the area of a rectangle is maximized when the rectangle is a square, the area of a triangle is maximized when the base and height are the same.

Here is a non-calculus approach.

Let x=base and y=height

$x+y=36\Rightarrow x=36-y$

$A=\frac{xy}{2}$

Make the sub and we get:

$A=\frac{x(36-x)}{2}$

$A=\frac{-x^{2}}{2}+18x$

Without calculus, the maximum can be found by using

$x=\frac{-b}{2a}$

$b=18, \;\ a=-1/2$

The base:

$x=\frac{-18}{2(-\frac{1}{2})}=18$

The height is then $36-18=18$