You are given a 12"x18" piece of construction paper. You are to cut a square out of each corner in order to create a 3-dimensional open-top box. What size squares would you need to cut in order to maximize the volume of the box?
let the side of the square to be cut out be x inches
resulting box is (12-2x) by (18-2x) by x
V = x(12-2x)(18-2x)= x(216 - 60x + 4x^2)
= 4x^3 - 60x^2 + 216x
dV/dx = 12x^2 - 120x + 216
= 0 for a max of V
x^2 - 10x + 18=0
solve for x
length = (18 - 2x)
width = (12 - 2x)
V = x (18-2x)(12-2x)
= x(4)(9-x)(6-x)
= 4 x (54-15x+x^2)
= 4 (54 x -15x^2 + x^3)
dV/dx = 0 for max
0 = 54 - 30 x + 3 x^2
0 = 18 - 10 x + x^2
x = [ 10 +/- sqrt (100-72) ]/2
[ 10 +/- 2 sqrt 7 ]/2
= 5 +/- sqrt 7
5+sqrt 7 is no good, more than half the width
so
5-sqrt 7 = 2.35
Looks like calculus not pre-calculus to me by the way. I do not see how to do it without taking the derivative.
what does sqrt mean?
To maximize the volume of the box, we need to find the dimensions that will result in the largest possible volume.
Let's start by visualizing the construction paper. The original dimensions are 12" by 18", and we need to cut squares out of each corner. Let's assume we cut x" by x" squares out of each corner.
By cutting out squares from each corner, the new dimensions of the construction paper will be:
Length: 12" - 2x"
Width: 18" - 2x"
To create a 3-dimensional open-top box, we will fold up the remaining flaps of the construction paper to form the sides of the box. The height of the box will be the length of the square we cut out.
Now, the volume of a box is calculated by multiplying its length, width, and height (V = lwh). In this case, the volume can be represented as:
V = (12 - 2x) * (18 - 2x) * x
To maximize the volume, we can use calculus to find the value of x that yields the maximum volume. By taking the derivative of the volume equation with respect to x and setting it equal to zero, we can solve for x.
Differentiating the volume equation:
dV/dx = -4x^3 + 60x^2 - 216x + 216
Setting dV/dx equal to zero:
-4x^3 + 60x^2 - 216x + 216 = 0
Solving this cubic equation might be complex, but we can use numerical methods or graphing calculators to find the root or approximate values of x where the derivative is zero. We can also use the Rational Root Theorem to identify potential rational solutions.
Once we find the value(s) of x that make the derivative equal to zero, we can substitute them back into the volume equation to get the corresponding maximum volume(s) for the box.
Note: If solving the cubic equation seems challenging, we can also try some trial and error by plugging in different values of x to see which one gives the highest volume.