A box with an open top is to be constructed from a square piece of cardboard, 10in wide, by cutting out a square from each other of the four and bending up the sides. What is the maximum volume of such a box?
Draw a square 10 by10 in.
Cut the squares x by x in each corner.
Notice the new dimension for the length and width of the base ( 10 - 2 x )
The area of the base = ( 10 - 2 x )²
To find the volume of a rectangular prism (your box), you must multiply the area of the base times the height.
V(x) = ( 10 - 2 x )² ∙ x
Now find first derivative.
V´(x) = [ ( 10 - 2 x )² ∙ x ]´
Use the product rule.
V´(x) = [ ( 10 - 2 x )² ∙ x ]´ = [ ( 10 - 2 x )² ]´ ∙ x + ( 10 - 2 x )² ∙ x´ =
[ 10² - 2 ∙ 10 ∙ 2 x + ( 2 x )² ]´ ∙ x + ( 10 - 2 x )² ∙ 1 =
( 100 - 40 x + 4 x² )´ ∙ x + ( 10 - 2 x )² =
( - 40 + 2 ∙ 4 x ) ∙ x + [ 10² - 2 ∙ 10 ∙ 2 x + ( 2 x)² ] =
( - 40 + 8 x ) ∙ x + 100 - 40 x + 4 x² =
- 40 x + 8 x² + 100 - 40 x + 4 x² =
12 x² - 80 x + 100
V´(x) = 12 x² - 80 x + 100
Second derivative:
V" (x) = [ V´(x) ]´ = [ 12 x² - 80 x + 100 ]´ = 12 ∙ 2 x - 80 = 24 x - 80
The function have extreme values ( maximum or minimum ) in points where first derivation = 0
In this case the function have extremes value in points where:
12 x² - 80 x + 100 = 0
The solutions are:
x = 5 / 3 and x = 5
Now you must do second derivative test.
If second derivative < 0 the function have maximum
If second derivative > 0 the function have minimum
In this case:
V" ( 5 / 3 ) = 24 ∙ 5 / 3 - 80 = 120 / 3 - 80 = 40 - 80 = - 40 < 0
V" ( 5 ) = 24 ∙ 5 - 80 = 120 - 80 = 40 > 0
So for x = 5 / 3 your function have maximum.
V max = V´( 5 / 3 ) = [ 10 - 2 ∙ ( 5 / 3 ) ] ² ∙ ( 5 / 3 ) =
( 10 - 10 / 3 ) ² ∙ ( 5 / 3 ) = ( 30 / 3 - 10 / 3 ) ² ∙ ( 5 / 3 ) =
( 20 / 3 ) ² ∙ ( 5 / 3 ) = 400 / 9 ∙ ( 5 / 3 ) = 2000 / 27
V max = 2000 / 27 in³ for x = 5 / 3 in
To find the maximum volume of the box with an open top, we need to determine the dimensions of the cut-out square and then calculate the volume.
Let's assume the side length of the cut-out square is x inches.
Since the original piece of cardboard is 10 inches wide, and we are cutting out squares from each corner, the width of the resulting box would be (10 - 2x) inches.
Similarly, the length of the box would also be (10 - 2x) inches as all sides are constructed by bending up the cardboard.
The height of the box would be x inches since we are bending up the cardboard at each corner.
Now, we can calculate the volume using the formula: Volume = Length x Width x Height.
Volume = (10 - 2x) * (10 - 2x) * x.
To find the maximum volume, we can take the derivative of the volume equation with respect to x and set it equal to zero:
dV/dx = 0.
Let's differentiate the volume equation:
dV/dx = 2(10 - 2x)(-2) * (10 - 2x) + (10 - 2x)(10 - 2x)(1) = 0.
Simplifying the equation, we get:
-4(10 - 2x)^2 + (10 - 2x)^2 = 0.
Now, let's solve for x:
-4(10 - 2x)^2 + (10 - 2x)^2 = 0.
Expanding and rearranging, we get:
(-40 + 8x + 40 - 8x + 1)x^2 = 0,
1x^2 = 0,
x^2 = 0.
From this, we can see that x = 0.
However, since x represents the side length of the cut-out square, it cannot be zero. Therefore, there's no valid solution for finding the maximum volume of the box.
Hence, the maximum volume of such a box cannot be determined.