+0  
 
0
83
1
avatar

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.

 Dec 6, 2020
 #1
avatar+114221 
+1

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

 

 

cool cool cool 

 Dec 6, 2020
edited by CPhill  Dec 6, 2020

26 Online Users

avatar
avatar