3. (15 points) A box without a top has dimensions a inches by y inches by z inch

Question

3. (15 points) A box without a top has dimensions a inches by y inches by z inches: The cost of the material for the base of the box is 4 cents per square inch. The cost of the material for the sides of the box is 1 cent per square inch. If the total budget is 48 cents, what are the dimensions of the box with the largest volume that can be produced? (You may assume without justification that only an absolute maximum exists. You may also assume that the dimensions at which this maximum occurs are all strictly positive.)

Explanation / Answer

No top

So, we have
area of the bottom is xy

Area of the sides is : 2yz + 2xz

So, cost of base is :
xy * 4
4xy cents

And cost of sides is :
1(2yz + 2xz)
= 2yz + 2xz cents

So, total cost is :
C = 4xy + 2yz + 2xz

Now, budget is 48 cents

So, 4xy + 2xz + 2yz = 48

Divide by 2 :
2xy + xz + yz = 24

And we gotta max xyz, the volume

So, we have :
f(x,y,z) = xyz ---> to be max-ed
g(x,y,z) : 2xy + xz + yz = 24

Now, we f ind partials :
fx = yz , fy xz , fz = xy
gx = 2y + z , gy = 2x + z , gz = x + y

Now, we link 'em like this
fx = m*gx , fy = m*gy and fz = m*gz...

yz = m(2y + z)
xz = m(2x + z)
xy = m(x + y)

Now, divide first two :
y/x = (2y+z)/(2x+z)

Crossmultiply :
2xy + yz = 2xy + xz
So, yz = xz
Thus x = y

Now, divide the second and third :
xz = m(2x + z)
xy = m(x + y)

Dividin' :
z/y = (2x+z )/(x+y)

Crossmultiply :
xz + yz = 2xy + yz
xz = 2xy
So, z = 2y

So, we have
x = y
y = y
z = 2y

And now from cost equation,
we have :
2xy + xz + yz = 24

2y^2 + 2y^2 + 2y^2 = 24
6y^2 = 24
y^2 = 4
y = 2

And thus x = 2

And z = 2y = 2*2 = 4

So, we have
x = 2
y = 2
z = 4

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