By second derivative test we see minima occurs at 3.39m .
Suppose that;
The length of one wire = x m
And the length of other wire is = (9-x) m
Length of the wires and bent into a equilateral triangle for a frame for a stained glass ornament respectively.
Area of each side of equilateral triangle length = [tex]\frac{X}{3 }[/tex] is [tex]\frac{\sqrt{3} }{4} ( \frac{x^{2} }{9} )[/tex] .
And that of circle given circumference ( 9−x ) is [tex]\frac{(9-x)^{2} }{4\pi }[/tex].
Total area = f (x) = [tex]\frac{\sqrt{3} }{4} (\frac{x^{2} }{9} ) + \frac{(9-x)^{2} }{4\pi }[/tex].
f'(x) = [tex]\frac{\sqrt{3} x}{4} + \frac{x-9}{2\pi }[/tex]
To minimize we differentiate, find the critical points and determine the local minima.
f′(x)= 0
we get ;
x = [tex]\frac{36}{2\sqrt{3}x +4 }[/tex]
x = [tex]= \frac{36}{10.87\\+4}\\\\\frac{36}{14.87}[/tex]
By second derivative test we see minima occurs at 3.39m .
for the more details about minima click the link given below.
https://brainly.in/question/41264339
