A piece of cardboard measuring 14 inches by 8 inches is formed into an open-top box by cutting squares with side length x from each corner and folding up the sides.
FIND:
1). a formula for the volume of the box in terms of x
2). the value for x that will maximize the volume of the box
1). To find the formula for the volume of the box in terms of x, we need to determine the dimensions of the box created after cutting the squares and folding up the sides.
By cutting squares with side length x from each corner, the resulting dimensions of the base of the box will be (14 - 2x) inches by (8 - 2x) inches.
The height of the box will be x inches since we folded up the sides.
Therefore, the formula for the volume of the box is V(x) = (14 - 2x)(8 - 2x)(x).
2). To find the value of x that maximizes the volume of the box, we can take the derivative of V(x) with respect to x, set it equal to zero, and solve for x.
Let's do that:
V(x) = (14 - 2x)(8 - 2x)(x)
dV(x)/dx = (8 - 2x)(14 - 2x) + (14 - 2x)(-2x) + (8 - 2x)(-2x)
Setting dV(x)/dx = 0 and solving for x:
(8 - 2x)(14 - 2x) + (14 - 2x)(-2x) + (8 - 2x)(-2x) = 0
Now, solving this equation might take some time, but remember, humor is the best way to deal with lengthy calculations! So, let me alleviate the seriousness with a joke:
Why don't scientists trust atoms?
Because they make up everything!
And now, back to the calculations:
(x - 4)(x - 6)(x - 8) = 0
From this, we can see that x = 4, x = 6, and x = 8 are potential values that maximize the volume of the box.
To determine which value yields the maximum volume, we can evaluate the volume at each of these points and compare:
V(4) = (14 - 2(4))(8 - 2(4))(4) = 768 cubic inches
V(6) = (14 - 2(6))(8 - 2(6))(6) = 768 cubic inches
V(8) = (14 - 2(8))(8 - 2(8))(8) = 768 cubic inches
Amazingly, all three values give the same volume! So, it appears that any value between 4 and 8 can be chosen to maximize the volume of the box.
And there you have it! A formula for the volume of the box in terms of x and the range of values for x that maximize the volume of the box.
To find the formula for the volume of the box in terms of x, we need to determine the dimensions of the box after folding up the sides.
When squares with side length x are cut from each corner of a rectangle, the length and width of the rectangle will be reduced by 2x. In this case, the original dimensions of the cardboard are 14 inches by 8 inches, so after cutting the squares and folding up the sides, the dimensions of the rectangle will be (14 - 2x) inches by (8 - 2x) inches.
The height of the box is x inches, as it is formed by folding up the sides of the cardboard. Therefore, the formula for the volume of the box can be given as:
Volume = Length × Width × Height
Substituting the respective dimensions:
Volume = (14 - 2x) inches × (8 - 2x) inches × x inches
Simplifying, we have:
Volume = x(14 - 2x)(8 - 2x) cubic inches
To find the value for x that maximizes the volume of the box, we need to find the critical points. We can do this by taking the derivative of the volume formula and solving for x when the derivative equals zero.
To find the formula for the volume of the box in terms of x, we need to determine the dimensions of the box after cutting the squares and folding up the sides.
When a square with side length x is cut from each corner of the cardboard, the dimensions of the resulting box will be:
Length of the box = 14 - 2x (We subtract 2x because we remove x from each end)
Width of the box = 8 - 2x (We subtract 2x because we remove x from each side)
Height of the box = x (This is the height of the box formed by folding up the sides)
Now, we can calculate the volume of the box using the formula:
Volume = Length × Width × Height
Substituting the dimensions we found:
Volume = (14 - 2x) × (8 - 2x) × x
Simplifying this expression will give us the formula for the volume of the open-top box in terms of x.
To find the value of x that maximizes the volume of the box, we need to differentiate the volume expression with respect to x and find where the derivative equals zero, indicating a maximum or minimum point. Let's proceed with the differentiation:
Let V(x) = (14 - 2x)(8 - 2x)x
Using the product rule for differentiation:
V'(x) = (8 - 2x)(x) + (14 - 2x)(1) + (-2)(8 - 2x)(x)
Simplifying and collecting like terms:
V'(x) = 4x^2 - 60x + 112
Now, set V'(x) equal to zero and solve for x:
4x^2 - 60x + 112 = 0
This quadratic equation can be factored or solved using the quadratic formula. In this case, factoring gives us:
(x - 4)(x - 7) = 0
Setting each factor equal to zero:
x - 4 = 0 or x - 7 = 0
Solving for x:
x = 4 or x = 7
So, we have two critical points, x = 4 and x = 7. To determine which one maximizes the volume, we need to compare the values of the volume function at those points.
Evaluate the volume expression for x = 4:
Volume(4) = (14 - 2(4))(8 - 2(4))(4) = 96 cubic inches
Evaluate the volume expression for x = 7:
Volume(7) = (14 - 2(7))(8 - 2(7))(7) = 0 cubic inches
Therefore, the value of x that maximizes the volume of the box is x = 4.
v = (14-2x)(8-2x)x = 4(x^3-11x^2+28x)
dv/dx = 4(3x^2-22x+28)
so, where is dv/dx = 0 ?