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?

Answer 1

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