Kendra will make a box with an open top by cutting, folding and taping a 12-inch by 16-inch rectangular piece of cardboard. She will begin by cutting from each corner of the flat cardboard a square with a side length that is a multiple of 0.5 inch. What is the maximum possible volume of the box? Express your answer as a decimal to the nearest tenth.
+128474 The base will have the dimensions ( 12 - 2* .5x) and ( 16 - 2 *.5x)
And the height will be .5x
Where x is the multiple we are looking for
V = (12 - 2 *.5x) (16 - 2*.5x) (.5x)
V = (12 - x) ( 16 - x) (.5x)
V = ( 192 - 28x + x^2) ( 1/2)x
V = ( 1/2)x^3 - 14x^2 + 96 x)
Take the derivative and set to 0
V'= (3/2)x^2 - 28x + 96 = 0
3x^2 - 56x + 192 = 0
This gives us some messy roots
Look at the graph here : https://www.desmos.com/calculator/imswwfyroh
The box has a max volume of ≈ 194.1 in^3 when x ≈ 4.5
