Question
A wire of length 4 m is cut in two parts and the first part is bent in the shape of a square, while the second part is bent in the shape of an equilateral triangle. For which lengths of these parts the total area enclosed by the square and the triangle is: a. minimized? b. maximized?
Answers
Steve
if the square has side s and the triangle has side t (both in cm), then
4s+3t = 400
a = s^2 + √3/4 t^2
Now just express t or s in terms of the other, and you have a quadratic for the area. The vertex of the parabola will give the minimum area.
The maximum will be realized when all of the wire is used for one shape.
4s+3t = 400
a = s^2 + √3/4 t^2
Now just express t or s in terms of the other, and you have a quadratic for the area. The vertex of the parabola will give the minimum area.
The maximum will be realized when all of the wire is used for one shape.