We want to construct a sample box for transporting medicine that should have a square base, and we only have 10m² of material to use in construction of the box- the box is supposed to have all six faces. Assuming that all the material is used in the construction process determine the maximum volume that the box can have.

Answer 1

Let's call x the length of one side of the square base and h the height of the box. Since the base is square, the area of the base is x². Each of the other four faces has an area of xh (since they are all rectangles with one side being x and the other being h). Finally, the top face also has an area of x².

We know that the total area of material we have is 10m², so we can write an equation:

x² + 4xh + x² = 10

Simplifying this equation, we get:

2x² + 4xh = 10

We want to maximize the volume of the box, which is given by V = x²h. To do this, we need to express h in terms of x and substitute it into the volume equation. From the equation above, we can solve for h:

h = (10 - 2x²) / (4x)

Now we can substitute this expression for h into the volume equation:

V = x²((10 - 2x²) / (4x))

Simplifying this expression, we get:

V = (5/2)x - (1/2)x³

To find the maximum volume, we need to find the critical points of this function. We can do this by taking the derivative and setting it equal to zero:

dV/dx = (5/2) - (3/2)x²

(5/2) - (3/2)x² = 0

5 - 3x² = 0

x = ±√(5/3)

We want the positive root for the length of the side of the base, since it must be a positive value. Substituting this value of x into the expression we found for h earlier, we get:

h = (10 - 2(5/3)) / (4√(5/3))

h = √(5/3)

Finally, we can substitute these values for x and h into the expression we found for the volume:

V = (5/2)√(5/3) - (1/2)(√(5/3))³

V ≈ 3.68m³

Therefore, the maximum volume that the box can have, given the amount of material available, is approximately 3.68m³.