Several optimization problems are solved and detailed solutions are presented. These problems involve optimizing functions in two variables using first and second order partial derivatives.
Solution to Problem 2:
Using all available cardboard to make the box, the total area A of all six faces of the prism is given by.
A = 2xy + 2yz + 2zx = 12
The volume V of the box is given by
V = xyz
Solve the equation 2xy + 2yz + 2zx = 12 for z
z = (6 - xy) / (x + y)
Substitute z in the expression of the volume V to obtain.
V(x,y) = xy (6 - xy) / (x + y)
Let us find the critical points by first finding the first order partial derivatives
Vx(x,y) = -y2( x2 + 2xy - 6 ) / (x + y) 2
Vy(x,y) = -x2( y2 + 2xy - 6 ) / (x + y) 2
We now solve the system of equations given by Vx = 0 and Vy = 0. One obvious solution is given by the point (0,0) but is not physically possible. Other solutions are found by setting
x2 + 2xy - 6 = 0
y2 + 2xy - 6 = 0
Subtracting the equations term by term we obtain
x2 - y2 = 0
Solve to obtain
x = y and x = - y
The solution x = -y is not valid for this problem since both x and y are dimensions and cannot be negative. Use x = y in the equation x2 + 2xy - 6 = 0, we obtain
x2 + 2x2 - 6 = 0
Solve for x
x = √2
y = x = √2
Let us now find the second order partial derivatives
Vxx(x,y) = -2y2( y2 + 6 ) / (x + y) 3
Vyy(x,y) = -2x2( x2 + 6 ) / (x + y) 3
Vxy(x,y) = -2xy(x2 + 3xy + y2 - 6 ) / (x + y) 3
We now need the values of Axx, Ayy and Axy to find the value of D = Vxx(√2,√2) Vyy(√2,√2) - Vxy2(√2,√2) in order to use the theorem on minima and maxima of functions with 2 variables.
Solution to Problem 3:
One way to find the distance from a point to a plane is to take a point (x,y,z) on the plane; find the distance between this point and the given point and minimize it. Because the distance involves the square root, it would be better to minimize the square of the distance. Let the square of the distance between the given point and the point (x,y,z) on the plane be f.
f(x,y,z) = (x - 1)2 + (y - 2)2 + (z + 1)2
We now solve the equation of the plane for z to obtain
z = 3 - x + y
Substitute z in f by 3 - x + y in f.
F(x,y) = (x - 1)2 + (y - 2)2 + (- x + y + 4)2
We now find the first order partial derivatives
Fx(x,y) = 2(x - 1) + 2(-1)(-x + y + 4)
Fy(x,y) = 2(y - 2) + 2(-x + y + 4)
We now need to find the critical points by setting the first partial derivatives equal to zero.
2(x - 1) + 2(-1)(-x + y + 4) = 0 and 2(y - 2) + 2(-x + y + 4) = 0
We now solve the system of equations to obtain
(8/3 , 1/3 , 2/3)
We now calculate the second order derivatives
Fxx(x,y) = 4
Fyy(x,y) = 4
Fxy(x,y) = -2
We now need to find the sign of D = Fxx(8/3,1/3) Fyy(8/3,1/3) - Fxy2(8/3,1/3) in order to use the theorem on minima and maxima of functions with 2 variables
D = 4 * 4 - (-2)2 = 12
Since D is positive and Fxx is positive, F has a minimum at the point (8/3,1/3) which correspond to a point on the plane given by
The distance d between the given point and the plane is given by
d = √ [ (1 - 8/3)2 + (2 - 1/3)2 + (-1 - 2/3)2 ]
= 5 / √3