A cardboard box without a lid is to have a volume of 8,788 cm3. Find the dimensions that minimize the amount of cardboard used.

x = y = 26 cm; z = 13 cm

Step-by-step explanation:

We can calculate the dimensions of the square base as

∛(2·8788) = 26 cm

the height of the box will be half of 26/2 which is 13 cm.

x = y = 26 cm; z = 13 cm

then the minimum area for the given volume can be calculated using what we call Lagrange multipliers, this makes it easier

area = xy +2(xz +yz)

But we were given the volume as 8788

Now we will make the partial derivatives of L to be in respect to the cordinates x, y, z, as well as λ to be equal to zero, then

L = xy +2(xz +yz) +λ(xyz -8788)

For x: we have

y+2z +λyz=0

For y we have

y: x +2z +λxz=0

For z we have 2x+2y +λxy=0eqn(*)

For we have xyz -8788=0

If we simplify the partial derivative equation of y and x above then we have

λ = (y +2z)/(yz).

= 1/z +2/yeqn(1)

λ = (x +2z)/(xz)

= 1/z +2/xeqn(2)

Set eqn(1 and 2) to equate we have

1/z +2/y = 1/z +2/x

x = y

From eqn(*) we can get z

λ = (2x +2y)/(xy) = 2/y +2/x

If we simplify we have

1/z +2y = 2/x +2/y

Then z = x/2

26/2 =13

Therefore,

x = y = 2z = ∛(2·8788)

X= 26

y = 26 cm

z = 13 cm

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.