Applications of Maxima and Minima Problems

Question

Find the lengths of the sides of an isosceles triangle with a given perimeter if its area is to be as large as possible. Find the lengths of the sides of an isosceles triangle with a given perimeter if its area is to be as large as possible.

Answer 1

Let P be the length of the perimeter..

Then if le let x be the length of one o the two equal sized sizes , then the length of the base of the

isosceles triangle y = P - 2 x.

Then the height of the triangle H = [ x² - ( P - 2 x )² /4 ] ^1/2

A ( x ) = (1/2) ( P - 2 x ) [ x² - ( P - 2 x )² /4 ] ^1/2 = (1/4) ( P - 2 x ) [4 x² - ( P - 2 x )² ] ^1/2.

A ( x ) = (1/4) ( P - 2 x ) [4 P x - P² ] ^1/2.

To maximize A ( x ) suffices to maximize Q (x ) = ( P - 2 x ) [ 4 P x - P²] ^1/2.

Then Q' (x ) ={ ( P - 2 x ) [4 P x - P² ] ^1/2.}'

=-2 [4 P x - P² ] ^1/2.+ (P - 2 x )[ 2P / [4 P x - P² ] ^1/2 ]

= { -2 [4 P x - P² ] + 2 P² - 4 P x } / { [4 P x - P² ] ^1/2 }

Q' (x ) = 0 ⇔ [-2(4 P x - P²) ] + 2 P² - 4 P x = 0 ⇔ -12 P x + 4 P² =0 12 P x = 4 P² x = P / 3

Q' (x ) > 0 if x < P / 3 and Q' (x ) < 0 if x > P / 3 Therefore at x = P / 3 Q (x ) attains maximum so A (x).

Notice that x = P / 3 .

Hence of all isosceles triangles with fixed perimeter the equilateral has the maximum area.

Answer 2

Let p: the fixed perimeter ; b: length of base ; x: length of congruent legs

b + 2x = p → x = (p - b) / 2

h² + (b/2)² = x² → h = √(4x² - b²) / 2

A = 1/2 bh = 1/2 b √(4[(p - b)/2]² - b²) / 2

A(b) = 1/4 b √(p² - 2pb) = √p/4 · [pb² - 2b³]^1/2

A^'(b) = √p/8 · (2pb - 6b²) · [pb² - 2b³]^-1/2 = 0

2pb = 6b²

p = 3b

Thus, predictably, an equilateral triangle will maximize area.