langrange multipliers?

Question

An open rectangular box without a top is to be constructed from material that costs $5 per square foot for the bottom and $2 per square foot for its sides. The bottom of the rectangular box is a square. Use the method of Lagrange multipliers (no credit will be given for any other method) to find the dimensions of the box of greatest volume that can be constructed for $240. You do not need to show that the answer you compute gives the greatest volume.

Answer 1

C = x^2*5 +2*4xy = 240
V= x^2*y maximum
F(x,y,L) = x^2*y+L(5*x^2+8xy-240)
dF/dx= 2*y*x +L(10x +8y)
dF/dy = x^2 +L(8x) L Lagrange multiplier
dF/dx=0
and
dF/dy= 0 from here x= 0 ( nonsense) or
x=-8L so L= -x/8
2x*y-x/8(10x+8y)=0
2x*y-5/4 x^2 -x*y=0 so xy=5/4 x^2Now going to the C equation
5x^2+8(5/4x^2) =240
x^2=16 and x=4 feet
80+32y =240 So y = 5 feet

Answer 2

You ask the greatest volume
V(x,y,z)=xyz
(x,y,z lengths where the box
is positioned with its sides
on x,y,z axis)
With the two restrictions
g(x,y,z)=0<->5xy+8xz-240=0
h(x,y,z)=0<->x-y=0
Must find n,m such as
Vf=nVg+mVh
yzi+xzj+xyk=n(5y+8z)i+n5xj+
+n8xk+mi-mj<-->
We take the system
yz=n(5y+8z)+m (2)
xz=5nx-m (3)
xy=8nx (4)
From(4) x=0 (not accept)
or y=8n
then x=8n and adding (2)+(3)
16nz=80n^2+8nz
hence 10n^2-nz=0
n=0(not accept)
z=10n
Sox=y=8n , z=10n
Now must
5(8n)^2+2*4*80n^2=240
n=1/2<->
x=y=4 and z=5

Answer 3

f(x,y,z) = sqrt [(x-a million)^2 + (y+a million)^2 + (z+a million)^2] It suffices to lessen [f(x,y,z)]^2 = (x-a million)^2 + (y+a million)^2 +(z+a million)^2, which i will call D(x,y,z). And the constrain is g(x,y,z) = x+4y+3z-2 = 0. proper the following ought to carry to boot to the constrain: D_x = (lambda)(g_x) D_y = (lambda)(g_y) D_z = (lambda)(g_z) D_x = 2(x-a million) = lambda D_y = 2(y+a million) = 4 lambda D_z = 2(z+a million) = 3 lambda. fixing for x, y and z, you get x = lambda/2 + a million, y = 2 lambda - a million, and z = 3(lambda)/2 - a million. Plugging those into the constrain g(x) = x+4y+3z-2 = 0, you get 13lambda - 8 = 0, so lambda = 8/13. using that to scrub up for x,y, and z, you get x = 17/13, y = 3/13 and z = -a million/13. putting those values decrease back into f(x,y,z), you need to get 4sqrt(2/13).

Answer 4

Let x = side of bottom of box
h = height of box
V = x^2h
A = x^2 + 4xh
C = $5A = $2.40
Let g(x,h) = 2.40 - 5(x^2 + 4xh)
Φ(x,h,λ) = x^2h + λ(2.4 - 5(x^2 + 4xh))
Φ(x,h,λ) = x^2h + λ(2.4 - 5x^2 - 20xh)
Φ(x,h,λ) = x^2h + 2.4λ - 5λx^2 - 20λxh
∂Φ/∂x = 2xh - 10λx - 20λh = 0 (1)
∂Φ/∂h = x^2 - 20λx = 0 (2)
∂Φ/∂λ = 2.4 - 5x^2 - 20λ = 0 (3)
From (1)
x = 20λ
From (2)
2.4 - 5x^2 - x = 0
5x^2 + x - 2.4 = 0
x = (-1 ± √(1 + 48))/10
x = (-1 ± 7)/10
Discarding the negative,
x = 0.3
λ = 0.3/20
From (1) & (2),
2(0.3)h - 10(0.3/20)(0.3) - 0.3h = 0
0.6h - 0.09/2 - 0.3h = 0
0.3h - 0.09/2 = 0
h = 0.3/2 = 0.15