Probem33+Elise,+Paul+F.

It would be nice to have a diagram here -- perhaps one of you can scan one in, or create one using a word (or even geometer's sketchpad) document, if you are able. I am assuming that x is the perimeter of the square and y is the perimeter of the equilateral triangle. By the way, this is a complicated problem and the values that I got were not so satisfying -- so I approximated them with a calculator. Also, don't forget to examine what happens at the endpoints! What is the minimum value for x? The maximum value?

x+y=10 y=10-x

A=[(1/4)x]^2 + (1/2)(y/3)h

(1/6 y)^2 + h^2 = (1/3 y)^2 h^2 = 4y^2/36 - y^2/36 h^2=y^2/12 h=(y/2)(1/3)^1/2

A=[(1/4)x]^2 + (1/2)(y/3)(y/2)(1/3)^1/2

optimize: A= (1/4)x]^2 + (1/2)((10-x)/3)((20-x)/2)(1/3)^1/2

Paul: I think what I did is a lot different from what you did. Here it is: You have a) a square with sides x and b) a triangle with sides b. Maximize: A = x² + .5bh To find the height h: b² = (b/2)² + h² h = sqrt(4b²/4 - b²/4) = sqrt(3b²/4) = sqrt(3)b/2 To find x in terms of b: 10 = 4x + 3b x = (10 – 3b)/4 A = (100 – 60b + 9b²)/16 + (.5b)(sqrt(3)b/2) A = 100/16 – 60b/16 + 9b²/16 + sqrt(3)b²/4 A = 6.25 – 3.75b + .9955b ² A’ = -3.75 + 1.991b = 0 3.75 = 1.991b b = 1.883 limit as b approaches 0 of A = 6.25 limit as b approaches 10/3 of A = 6.25 – (3.75)(10/3) + (.9955)(100/9) = 4.811 limit as b approaches 1.883 of A = 6.25 – (3.75)(1.833) + (.9955)(1.833)² = 2.718 Therefore, the area is maximized as the dimensions of the triangle approach 0 m X 0 m X 0 m and the dimensions of the square approach 2.25 m X 2.25 m X 2.25 m X 2.25 m. The area is minimized as the dimensions of the triangle approach (10/3) m X (10/3) m X (10/3) m and the dimensions of the square approach 0 m X 0 m X 0 m X 0 m.