Optimization-A rod with isosceles-triangular cross section

[wind]

A rod with isosceles-triangular cross section is machined from a cylindrical steel rod of diameter 2cm. The strength of the rod is directly proportional to the product of the base of the triangle and the square of the height of the triangle. What are the dimensions of such a rod that has maximum strength?

let s rep strenght

dh^2=s

I don't understand this question, what equations am I supposed to use?

 

[skeeter]

Re: Optimization-A rod with isosceles-triangular cross secti

A rod with isosceles-triangular cross section is machined from a cylindrical steel rod of diameter 2cm. The strength of the rod is directly proportional to the product of the base of the triangle and the square of the height of the triangle. What are the dimensions of such a rod that has maximum strength?

let s rep strenght

dh^2=s read the bold type again ... the equation for strength should be s = kbh², where k is a constant of proportionality.

you know that the isosceles triangle will fit inside a circle of radius 1 cm since it was machined from a cylinder.

sketch a picture of a circle of radius 1 cm centered at the origin. let the base of the isosceles triangle be in quadrants I and II parallel to the x-axis, and the vertex of the isosceles triangle be at the point (0,-1).

base of the triangle is 2x

height of the triangle is 1 + sqrt(1 - x²)

s = k*2x*[1 + sqrt(1 - x²)]²

find ds/dx and maximize. note that 0 < x < 1.