we want to construct a box with a square base and no top. the volume of the box will be 8 cm^3 what are the dimension of the minimize the surface area .
let base be x by x
let the height be y
x^2 y = 8
y = 8/x^2
SA = base + 4 sides
= x^2 + 4xy
= x^2 + 4(8/x^2)
= x^2 + 32/x
d(SA)/dx = 2x - 32/x^2
= 0 for a minimum of SA
2x = 32/x^2
2x^3 = 32
x^3 = 16
x = cuberoot(16) = appr 2.52
then y = 1.26
the base is 2.52 by 2.52 and the height is 126
(notice the height is half of the side of the base)
To minimize the surface area of the box with a square base and no top while keeping the volume constant, we need to find the dimensions of the box.
Let's start by defining the variables:
- Let's say the length of one side of the square base is "x" cm.
- Since the box has no top, the height will also be "x" cm.
Now let's calculate the dimensions that will minimize the surface area.
1. Calculate the volume of the box:
V = base area × height
V = x^2 × x
V = x^3
Since we know the volume is 8 cm^3, we can set up an equation:
x^3 = 8
2. Solve for x:
By taking the cube root of both sides, we can find the value of x.
x = ∛(8)
x = 2 cm
3. Calculate the surface area:
The surface area of the box is the sum of the areas of all its faces.
Surface area = 2 × (base area) + 4 × (base side × height)
Substituting the values:
Surface area = 2 × (x^2) + 4 × (x × x)
Surface area = 2x^2 + 4x^2
Surface area = 6x^2
Substitute x = 2 cm:
Surface area = 6(2)^2
Surface area = 6 × 4
Surface area = 24 cm^2
Therefore, to minimize the surface area of the box, the dimensions should be 2 cm for each side of the square base and 2 cm for the height.