A 22-inch by 70-inch piece of cardboard is used to make an open-top box by removing a square from each corner of the cardboard and folding up the flaps on each side. What size square should be cut from each corner to get a box with the maximum volume

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Answer:

5 Inches

Step-by-step explanation:

Let the side length of the square in inches=y.

Height of the Box = y

Since we are removing the square from both sides

  • Width of the box = 22-2y
  • Length of the box = 70-2y

Then the volume of the box,  

[tex]V = y(22 -2y)(70-2y) \\= 4y(y -11)(y -35)\\V = 4y^3 -184y^2 +1540y[/tex]

We maximize the volume of the box by taking its derivative and solving it for its critical point.

[tex]V' = 12y^2 -368y +1540[/tex]

When V'=0

[tex]12y^2 -368y +1540=0[/tex]

Divide all through by 4 and factor

[tex](3x -77)(x -5) = 0 \\3x-77=0$ or $x-5=0\\x=25.6$ or x=5[/tex]

x cannot be greater than 11. therefore, the value of x=25.6 is extraneous. The side length of the square that should be cut to maximize its volume is 5 inches.