Question

g(x,y)= 2x^2+3y^2

subject to: 2x+2y<_1

Answer #1

Solve the IVP using the Eigenvalue method.
x'=2x-3y+1
y'=x-2y+1
x(0)=0
y(0)=1
x'=2x-3y+1
y'=x-2y+1
x(0)=0
y(0)=1
Solve
the IVP using the Eigenvalue method.
x'=2x-3y+1
y'=x-2y+1
x(0)=0
y(0)=1

Solve by using a matrix exponential
x' = x - 2y + 2e-t
y' = 2x-3y

Use
Gaussian Elimination to solve and show all steps:
1. (x+4y=6)
(1/2x+1/3y=1/2)
2. (x-2y+3z=7)
(-3x+y+2z=-5)
(2x+2y+z=3)

use lagrange multipliers to locate the maximum of f(x,y,z) =
2x^2 - 2y + z^2 subject to the constraint x^2 + y^2 + z^2 = 1

Maximize the function: ln(x) + y subject to 2x + 3y = 12 using
the Lagrange method where ln(x) is the (natural) log of x

Consider the following linear system:
x + 2y + 3z = 6
2x - 3y + 2z = 14
3x + y - z = -2
Use Gaussian Elimination with Partial Pivoting to
solve a solution in an approximated sense.

if
g(x,y)= (x^2)+(3y^2)-2x
a) what is the only critical point
b) at the critical point, does g have a local minimum, local
maximum, or a saddle point?

find d^2y/dx^2 for 3y^2-xy+2x^2=2

Solve by eliminating the arbitrary functions:
U = f (y-2x) + g (2y-x)

Minimize C = −2x + 3y subject to 3x + 4y ≤ 24, 7x − 4y ≤ 16, and
x, y ≥ 0.

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