MATH SOLVE

2 months ago

Q:
# You are constructing a box from a piece of cardboard with dimensions 6 by 8 meters. You cut equal-size squares from each corner of the cardboard so you may fold the edges to construct the open top box. What are the dimensions of the box with the largest volume?

Accepted Solution

A:

Answer:Length = 3.74 metersWidth = 5.74 metersHeight = 1.13 meters Step-by-step explanation:Dimensions of the given piece of cardboard is 6 by 8 meters.squares from each corner of the cardboard.Then the dimensions of the box formed when folded Length = (6 - 2x) metersWidth = (8 - 2x) metersHeight = x metersVolume of the box = Length × width × heightVolume (V) = (6 - 2x)(8 - 2x)xV = x(48 - 12x - 16x + 4x²)V = 4x³ - 28x² + 48xWe take the derivative of Volume V[tex]\frac{dV}{dx}=\frac{d}{dx}(4x^{3}-28x^{2}+48x)[/tex][tex]\frac{dV}{dx}=12x^{2}-56x+48[/tex]For maximum value,[tex]\frac{dV}{dx}=12x^{2}-56x+48=0[/tex]3x² - 14x + 12 = 0[tex]x=\frac{14\pm \sqrt{(14)^{2}-4(3)(12)}}{6}[/tex][tex]x=\frac{14\pm \sqrt{196-144}}{6}[/tex][tex]x=\frac{14\pm 7.2}{6}[/tex]x = 3.53, 1.13Now we have to check the value of x at which volume is maximum.We take the second derivative of VV" = 24x - 56At x = 3.53V" = 24(3.53) - 56 = 28.72 Since the value of V" is positive so the volume is not maximum at x = 3.53At x = 1.13V" = 24(1.13) - 56V" = 27.2 - 56 = - 28.8 < 0 Therefore, for x = 1.13, volume will be maximum.Now length of the box = 6 - 2(1.13) = 3.74 meterswidth = 8 - 2(1.13) = 5.74 metersAnd height of the box = 1.13 meters.