+97 A piece of wire of length 22 m is to be cut into two pieces. One piece will be used to make a square and the other to make an equilateral triangle. What are the two lengths of wire required such that the total area enclosed is a minimum?
Length of triangle string?
Length of rectangle string?
Thanks again guys, also can you please show some working out, thanks so much.
+128577 Let 22-x be the length reserved for the square and let (x) be the length reserved for the triangle.
So (22-x)/4 is be length of the side of the square and (x/3) is the length of the side of the equilateral triangle.
So the area of the square is given by [(22-x)/4]^2 = (1/16)[22-x]^2. And the area of the triangle is given by
(1/2)(x/3)^2sin(60) = (1/2)(1/9)x^2(√3/2) = (√3/36)x^2
So the total area is given by
A = (1/16)[22-x]^2 + (√3/36)x^2
Instead of doing the calculus...we can graph this function and find where the area is minimized. Here is the graph: https://www.desmos.com/calculator/jyd2p81j5z
So, it appears that the area is minimized when about x = 12.4 m is reserved for the triangle and (22-x) = 9.6m is reserved for the square.
Make the square 1 m each side. Area= 1X1=1 sq. m Triangle is 18/3= 6m each side. Area of triangle = (6X6)/2=18 18plus 1= 19 sq.m
+128577 Let 22-x be the length reserved for the square and let (x) be the length reserved for the triangle.
So (22-x)/4 is be length of the side of the square and (x/3) is the length of the side of the equilateral triangle.
So the area of the square is given by [(22-x)/4]^2 = (1/16)[22-x]^2. And the area of the triangle is given by
(1/2)(x/3)^2sin(60) = (1/2)(1/9)x^2(√3/2) = (√3/36)x^2
So the total area is given by
A = (1/16)[22-x]^2 + (√3/36)x^2
Instead of doing the calculus...we can graph this function and find where the area is minimized. Here is the graph: https://www.desmos.com/calculator/jyd2p81j5z
So, it appears that the area is minimized when about x = 12.4 m is reserved for the triangle and (22-x) = 9.6m is reserved for the square.
