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.
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.
To find the maximum volume that the box can have, we need to optimize the dimensions of the box.
Let's assume that the side length of the square base is x meters. We also need to account for the height of the box, which we'll call h meters.
The surface area of the box is given by the sum of the areas of all six faces. In this case, it's equal to the given area of 10m².
Surface area of the box = 2(squared base) + 4(height * width) = 2(x²) + 4(xh)
Since we want to maximize the volume, we can use the constraint that all the material must be used in the construction process. Thus, the total surface area of the box must be equal to 10m².
2(x²) + 4(xh) = 10
We can rearrange this equation to express the height h in terms of the side length x:
h = (10 - 2(x²)) / (4x)
Now we can express the volume V of the box in terms of x and h:
V = (base area) * height
V = (x²) * h
V = (x²) * (10 - 2(x²)) / (4x)
V = (10x - 2(x⁴)) / 4
To find the maximum volume, we need to find the critical points by taking the derivative of V with respect to x and set it equal to zero:
dV/dx = 0
10/4 - 8(x³) / 4 = 0
10 - 8(x³) = 0
10 = 8(x³)
x³ = 10/8
x³ = 5/4
x = (5/4)^(1/3)
x ≈ 1.1487
To determine if this critical point is a maximum, we can take the second derivative of V with respect to x:
d²V/dx² = -24(x²) / 4
d²V/dx² = -6x²
Substituting x = (5/4)^(1/3) into the second derivative, we find:
d²V/dx² = -6((5/4)^(1/3))²
d²V/dx² = -6(5/4)^(2/3)
d²V/dx² ≈ -4.2945
The second derivative is negative, which suggests that the critical point x ≈ 1.1487 is a local maximum.
To find the corresponding height h, we can substitute x into the equation we derived earlier:
h = (10 - 2(x²)) / (4x)
h ≈ (10 - 2((5/4)^(1/3))²) / (4(5/4)^(1/3))
h ≈ (10 - 2(5/4)^(2/3)) / (4(5/4)^(1/3))
h ≈ (10 - 2(5/4)^(2/3)) / (4(5/4)^(1/3))
Now, we can calculate the maximum volume by substituting the values of x and h into the volume equation:
V = (10x - 2(x⁴)) / 4
V = (10 * (5/4)^(1/3) - 2((5/4)^(1/3))⁴) / 4
By evaluating this expression, we can find the maximum volume that the box can have.