Sec2Sess41

Unformatted notes below. Notes on my shorthand method

Session 41: Advanced Example

Want to build a pyramid with a given triangular base and given volume

Where to put the top of the pyramid is the method.

Goal is to minimize the total surface area.

Vol = 1/3*area(base)*height

In this problem the height is fixed

The problem could be described as: P1 = (x1,y1,0), P2 = (x2,y2,0), P3 = (x3,y3,0), P = (x,y,h) where P1,P2,P3 are points on the triangle sides and P is the point on the top.

Finding the solution is too complicated for directly using cross product formulas.

Pyramid Area = sum of 3 triangle bases * 3 heights

Q (x,y,0) is center point of bottom triangle in pyramid

u1, u2, and u3 are distances from Q to the center of the sides of the bottom triangle

Area of triangle with points P, Q, and u1 is sqrt(u1^2+h^2)

these triangles are "face triangles"

Side area = 1/2*a1*(u1^2+h^2) + 1/2*a2*(u2^2+h^2) + 1/2*a3*(u3^2+h^2)

a function of 3 variables = f(u1,u2,u3)

cut base into 3 => Area(base) = 1/2*ai*u1+1/2*a2*u2+1/2*a3*u3

for lagrange multiplications g/(Area(base))

grad(f) = lamda*grad(g): pd(f)/pd(u1)=1/2*a1*u1/sqrt(u1^2+h^2) = lamda*1/2*a1

1/2*a1 cancels out on both sides of the equation leaving

pd(f)/pd(u1)=u1/sqrt(u1^2+h^2) = lamda

equations are the same for u2, u3, etc. Because all equations equal lamda then u1 = u2 = u3

Therefore Q is equidistant from all sides and is named the "incenter"

Example Problem:

Find the maximum and minimum values of the function f(x,y,z) = x^2+x+2y^2+3z^2 as (x,y,z) varies on the unit sphere x^2+y^2+z^2=1

Method: find points where the gradient of the objective function is parallel to the constraint function.

pd(x) = 2x+1=lamda*2x

pd(y) = 4y = lamda*2y

pd(z) = 6z = lamda*2z

x^2+y^2+z^2=1

2cnd equ. (for y): for result of 0: y=0 or lamda = 2

3rd equ. (for z): z=0 or lamda=3

Solutions:

a) y=z=0 => x=1 or x=-1 (1,0,0),(-1,0,0)

b) y=0,lamda=3 => x=1/4,z=+/-sqrt(15)/4 creates (1/4,0,sqrt(15)/4),(1/4,0,-sqrt(15)/4)

c) lamda=2,z=0 => x = 1/2, y = +/- sqrt(3)/2 creates (1/2,sqrt(3)/2,0),(1/2,-sqrt(3)/2,0)

applying the solution points shows where the function f is maximized

a) creates function values 2 or 0, b) 25/8 c) 9/4

looking at the value the min+max can be found. min value of f is 0 and 25/8 is the max.

overall method: solve system of equations from partial derivatives to find solutions.

Use solutions in function to find min/max result values.