A cardboard box of 108in cubed volume with a square base and no top constructed. Find the minimum area of the cardboard needed. (Optimization)
Would the answer be 3.7?
you are looking for area, finding s is only the first step. Look at the area formula.
answer in the back of the book is 180in squared
but im not getting that answer
To find the minimum area of the cardboard needed, we'll need to optimize the design of the box.
Let's start by identifying the variables involved in this problem:
- Side length of the square base (let's call it x)
- Height of the box (let's call it h)
To proceed with optimization, we can use the given information that the volume of the box is 108 in³. The volume of the box can be calculated as the product of the base area and the height:
Volume = Base area * Height
108 = x² * h
Now, let's express the surface area of the cardboard in terms of x and h. A box with no top has five sides: the base, four vertical sides, and no top. So, the cardboard surface area can be expressed as:
Surface Area = Base Area + 4 * Side Area
Surface Area = x² + 4 * (x * h)
We need to express the surface area in terms of a single variable to optimize it. Since we have the volume equation expressed in terms of h, we can solve it for h and substitute the value in the surface area equation:
108 = x² * h
h = 108 / x²
Surface Area = x² + 4 * (x * h)
Surface Area = x² + 4 * (x * (108 / x²))
Now, simplify the equation:
Surface Area = x² + 432 / x
To minimize the surface area, we need to find the critical points, where the derivative is equal to zero. Let's differentiate the surface area equation:
d(Surface Area) / dx = 2x - 432 / x²
To find the critical points, set the derivative equal to zero and solve for x:
2x - 432 / x² = 0
Multiplying through by x²:
2x³ - 432 = 0
Now, solve for x:
2x³ = 432
x³ = 216
x = 6
So, the critical point is x = 6. Since we are dealing with a physical box, the base length cannot be negative, so x = 6 is the only critical point.
Thus, the minimum area of the cardboard needed for the box is given by plugging the value of x back into the surface area equation:
Surface Area = x² + 432 / x
Surface Area = 6² + 432 / 6
Surface Area = 36 + 72
Surface Area = 108 square inches
Therefore, the minimum area of the cardboard needed to construct the box is 108 square inches.
area= s^2+hs
Volume=s^2*h or 108/s^2=h
area= s^2+108/s
dA/ds= 2s-108/s^2=0
2s^3=108
solve.