#Irina Mukhametzhanova, Section 24
#This homework is OK to post

#14.8 Homework
#Exercise 5:
#First, we find the gradients of both of the functions:
#grad(f) = <f_x,f_y> = <2*x, 2*y>
#grad(g) = <g_x,g_y> = <2,3>
#We then introduce a constant L that makes grad(f) a multiple of
#grad(g), like so:
#grad(f) = L*grad(f)
#<2*x, 2*y> = L*<2,3>
#We have a system of equations:
#Eq. 1: 2*x = 2*L
#Eq. 2: 2*y = 3*L
#Eq. 3: 2*x + 3*y = 6
#Because we have 3 equations and 3 unknowns, we can solve for L, x, and y:
#(2*y = 3*L) * (3/2)
#3*y = (9/2)*L
#Plug into Eq. 3:
#2*x + 3*y = 6
#2*L + (9/2)*L = 6
#(13/2)*L = 6
#13*L = 12
#L = 12/13
#Solve for x:
#2*x = 2*L
#x = L
#x = 12/13
#Solve for y:
#2*y = 3*L
#y = (3/2)*(12/13)
#y = 18/13
#So, our critical point is (12/13, 18/13)
#We plug the point into the original function:
#f(12/13, 18/13) = (12/13)^2 + (18/13)^2 = (144 + 324)/(169) = 
# = 468/169 = 36/13
#ANSWER: The minimum of this function with the constraint is 36/13,
#and does not have a maximum

#Exercise 7:
#First, we find the gradients of both of the functions:
#grad(f) = <f_x,f_y> = <y, x>
#grad(g) = <g_x,g_y> = <8*x, 18*y>
#We then introduce a constant L that makes grad(f) a multiple of
#grad(g), like so:
#grad(f) = L*grad(f)
#<y, x> = L*<8*x, 18*y>
#We have a system of equations:
#Eq. 1: y = 8*L*x
#Eq. 2: x = 18*L*y
#Eq. 3: 4*x^2 + 9*y^2 = 32
#Because we have 3 equations and 3 unknowns, we can solve for L, x, and y:
#Plug Eq. 2 into Eq. 1:
#y = 8*L*18*L*y
#144*L^2 = 1
#L^2 = 1/144
#L = +-1/12
#Plug L=1/12 into Eq. 2:
#x = 18/12*y
#x = (3/2)*y
#Square this equation and plug x^2 into Eq. 3:
#x^2 = (9/4)*y^2
#4*(9/4)*y^2 + 9*y^2 = 32
#9*y^2 + 9*y^2 = 32
#18*y^2 = 32
#y^2 = 32/18
#y = +-4/3
#x = +-2
#Now, plug L=-1/12 into Eq. 2:
#x = 18/(-12)*y
#x = (-3/2)*y
#Square this equation and plug x^2 into Eq. 3:
#x^2 = (9/4)*y^2
#4*(9/4)*y^2 + 9*y^2 = 32
#9*y^2 + 9*y^2 = 32
#18*y^2 = 32
#y^2 = 32/18
#y = +-4/3
#x = -+2
#So, our critical points are (4/3,2), (-4/3,2), (4/3,-2) and (-4/3,-2)
#Plug each into the function:
#f(4/3,2) = 8/3
#f(-4/3,2) = -8/3
#f(4/3,-2) = -8/3
#f(-4/3,-2) = 8/3
#ANSWER:The maximum value of the function with the constraint is 8/3,
#and the minimum value is -8/3

#Exercise 9:
#First, we find the gradients of both of the functions:
#grad(f) = <f_x,f_y> = <2*x, 2*y>
#grad(g) = <g_x,g_y> = <4*x^3, 4*y^3>
#We then introduce a constant L that makes grad(f) a multiple of
#grad(g), like so:
#grad(f) = L*grad(f)
#<2*x, 2*y> = L*<4*x^3, 4*y^3>
#We have a system of equations:
#Eq. 1: 2*x = 4*L*x^3
#Eq. 2: 2*y = 4*L*y^3
#Eq. 3: x^4 + y^4 = 1
#Because we have 3 equations and 3 unknowns, we can solve for L, x, and y:
#We can already see that either x or y are 0, but not both.
#If x=0, y = 1 and -1
#If y=1, x = 1 and -1
#Looking at Eq. 1:
#2*x = 4*L*x^3
#x = 2*L*x^3
#1/(2*L) = x^2
#x = sqrt(1/(2*L))
#Same operations for y:
#y = sqrt(1/(2*L))
#Plug them into the Eq. 3:
#(sqrt(1/(2*L)))^4 + (sqrt(1/(2*L)))^4 = 1
#1/(4*L^2) + 1/(4*L^2) = 1
#2/(4*L^2) = 1
#2 = 4*L^2
#L^2 = 1/2
#L = +- sqrt(1/2)
#Plug both values into the equation for x:
#x = sqrt(1/(2*sqrt(1/2))) = 1/(2^(1/4))
#x = sqrt(1/(-2*sqrt(1/2))) = -1/(2^(1/4))
#And y:
#x = sqrt(1/(2*sqrt(1/2))) = 1/(2^(1/4))
#x = sqrt(1/(-2*sqrt(1/2))) = -1/(2^(1/4))
#So, our points are (+-1/(2^(1/4)), +-1/(2^(1/4))), (+-1,0), and (0,+-1)
#We plug in each point into the function:
#f(+-1/(2^(1/4)), +-1/(2^(1/4))) = (1/(2^(1/4))^2 + (1/(2^(1/4))^2 = 
# = 1/sqrt(2) + 1/sqrt(2) = 2/sqrt(2) = sqrt(2)
#f(+-1,0) = 1
#f(0,+-1) = 1
#ANSWER: The maximum value of the function with the constraint is sqrt(2),
#and the minimum value is 1

#Exercise 11:
#First, we find the gradients of both of the functions:
#grad(f) = <f_x,f_y,f_z> = <3,2,4>
#grad(g) = <g_x,g_y,g_z> = <2*x,4*y,12*z>
#We then introduce a constant L that makes grad(f) a multiple of
#grad(g), like so:
#grad(f) = L*grad(f)
#<3,2,4> = L*<2*x,4*y,12*z>
#We have a system of equations:
#Eq. 1: 2*L*x = 3
#Eq. 2: 4*L*y = 2
#Eq. 3: 12*L*z = 4
#Eq. 4: x^2 + 2*y^2 + 6*z^2 = 1
#Because we have 4 equations and 4 unknowns, we can solve for L, x, y, and z:
#Square Eq. 1:
#(2*L*x = 3)^2
#4*L^2*x^2 = 9
#x^2 = 9/(4*L^2)
#Square Eq. 2:
#(4*L*y = 2)^2
#16*L^2*y^2 = 4
#y^2 = 1/(4*L^2)
#Square Eq. 3:
#(12*L*z = 4)^2
#144*L^2*z^2 = 16
#z^2 = 1/(9*L^2)
#Plug the values for x^2, y^2, and z^2 into Eq. 4:
#x^2 + 2*y^2 + 6*z^2 = 1
#9/(4*L^2) + 2*(1/(4*L^2)) + 6*(1/(9*L^2)) = 1
#Multiply both sides by 36:
#81/L^2 + 18/L^2 + 24/L^2 = 36
#(81 + 18 + 24) = 36*L^2
#123 = 36*L^2
#L = sqrt(123/36)
#Plug L into the equations for x^2, y^2, and z^2
#x^2 = 9/(4*(sqrt(123/36))^2)
#x^2 = 9/(4*(123/36))
#x^2 = 9/(123/9)
#x^2 = 81/123
#x = +-9/sqrt(123)
#y^2 = 1/(4*(sqrt(123/36))^2)
#y^2 = 1/(4*(123/36))
#y^2 = 1/(123/9)
#y^2 = 9/123
#y = +-3/sqrt(123)
#z^2 = 1/(9*(sqrt(123/36))^2)
#z^2 = 1/(9*(123/36))
#z^2 = 1/(123/4)
#z^2 = 4/123
#z = +-2/sqrt(123)
#So, our possible points are (+-9/sqrt(123), +-3/sqrt(123), +-2/sqrt(123))
#Because f(x,y,z) is just adding the three variables together, the maximum
#will occur when all three are positive, and the minimum will occur when
#all three are negative:
#f(+9/sqrt(123), +3/sqrt(123), +2/sqrt(123)) = 
# = 3*9/sqrt(123) + 2*3/sqrt(123) + 4*2/sqrt(123) = 
# = (27 + 6 + 8)/sqrt(123) = 41/sqrt(123)
#f(-9/sqrt(123), -3/sqrt(123), -2/sqrt(123)) = 
# = -3*9/sqrt(123) - 2*3/sqrt(123) - 4*2/sqrt(123) = 
# = (-27 - 6 - 8)/sqrt(123) = -41/sqrt(123)
#ANSWER: The maximum value of the function with the constraint is 41/sqrt(123),
#and the minimum value is -41/sqrt(123)

#Exercise 13:
#First, we find the gradients of both of the functions:
#grad(f) = <f_x,f_y,f_z> = <y,x,2>
#grad(g) = <g_x,g_y,g_z> = <2*x,2*y,2*z>
#We then introduce a constant L that makes grad(f) a multiple of
#grad(g), like so:
#grad(f) = L*grad(f)
#<y,x,2> = L*<2*x,2*y,2*z>
#We have a system of equations:
#Eq. 1: y = 2*L*x
#Eq. 2: x = 2*L*y
#Eq. 3: 2 = 2*L*z
#Eq. 4: x^2 + y^2 + z^2 = 36
#Because we have 4 equations and 4 unknowns, we can solve for L, x, y, and z:
#Plug Eq. 2 into Eq. 1:
#y = 2*L*2*L*y
#4*L^2 = 1
#L^2 = 1/4
#L = +-(1/2)
#First, plug L=1/2 into each equation:
#y = x
#x = y
#z = 2
#Plug y=x and z=2 into Eq. 4 and solve for x:
#x^2 + x^2 + 2^2 = 36
#2*x^2 = 32
#x = +-4
#y = +-4
#z = 2
#So, some of our points are (4,4,2) and (-4,-4,2)
#Now, plug in L=-(1/2) instead:
#y = -x
#x = -y
#z = -2
#Plug y=-x and z=2 into Eq. 4 and solve for x:
#x^2 + x^2 + 2^2 = 36
#2*x^2 = 32
#x = +-4
#y = -+4
#z = -2
#So, our critical points are (4,4,2), (-4,4,-2), (4,-4,-2) and (-4,-4,2)
#Plug each of them into the function:
#f(4,4,2) = 4*4 + 2*2 = 16 + 4 = 20
#f(-4,4,-2) = (-4)*(4) + 2*(-2) = -16 - 4 = -20
#f(4,-4,-2) = (4)*(-4) + 2*(-2) = -16 - 4 = -20
#f(-4,-4,2) = (-4)*(-4) + 2*2 = 16 + 4 = 20
#ANSWER: The maximum value of the function with the constraint is 20,
#and the minimum value is -20

#Exercise 15:
#First, we find the gradients of both of the functions:
#grad(f) = <f_x,f_y,f_z> = <y+z,x,x>
#grad(g) = <g_x,g_y,g_z> = <2*x,2*y,2*z>
#We then introduce a constant L that makes grad(f) a multiple of
#grad(g), like so:
#grad(f) = L*grad(f)
#<y+z,x,x> = L*<2*x,2*y,2*z>
#We have a system of equations:
#Eq. 1: y+z= 2*L*x
#Eq. 2: x = 2*L*y
#Eq. 3: x = 2*L*z
#Eq. 4: x^2 + y^2 + z^2 = 4
#Because we have 4 equations and 4 unknowns, we can solve for L, x, y, and z:
#Eq. 2 and Eq. 3 are equal:
#2*L*y = 2*L*z
#y = z
#Go back to Eq. 1:
#y+z= 2*L*x
#y+y = 2*L*2*L*y
#2*y = 4*L^2*y
#4*L^2 = 2
#L^2 = 1/2
#L = +-sqrt(1/2)
#Go back to equations 2 and 3:
#y = x/(2*L)
#y = x/(2*sqrt(1/2))
#z = x/(2*L)
#z = x/(2*sqrt(1/2))
#Plug them into Eq. 4:
#x^2 + (x/(2*sqrt(1/2)))^2 + (x/(2*sqrt(1/2)))^2 = 4
#x^2 + x^2/2 + x^2/2 = 4
#2*x^2 = 4
#x^2 = 2
#x = +-sqrt(2)
#y = +-sqrt(2)/(2*sqrt(1/2))
#y = +-1
#z = +-1
#Now, do this with L=-sqrt(1/2):
#y = x/(2*L)
#y = x/(-2*sqrt(1/2))
#z = x/(2*L)
#z = x/(-2*sqrt(1/2))
#Plug them into Eq. 4:
#x^2 + (x/(-2*sqrt(1/2)))^2 + (x/(-2*sqrt(1/2)))^2 = 4
#x^2 + x^2/2 + x^2/2 = 4
#2*x^2 = 4
#x^2 = 2
#x = +-sqrt(2)
#y = +-sqrt(2)/(-2*sqrt(1/2))
#y = -+1
#z = -+1
#So, our possible critical points (sqrt(2), 1, 1), (sqrt(2), -1, -1), 
#(-sqrt(2), 1, 1), and (-sqrt(2), -1, -1)
#Plug them both into the function:
#f(sqrt(2), 1, 1) = sqrt(2) + sqrt(2) = 2*sqrt(2)
#f(sqrt(2), -1, -1) = -sqrt(2) - sqrt(2) = -2*sqrt(2)
#f(-sqrt(2), 1, 1) = -sqrt(2) - sqrt(2) = -2*sqrt(2)
#f(-sqrt(2), -1, -1) = sqrt(2) + sqrt(2) = 2*sqrt(2)
#ANSWER: The maximum value of the function with the constraint is 2*sqrt(2),
#and the minimum value is -2*sqrt(2)