A box with an open to is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides. find the largest volume that such a box can have?

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Answer 1 · 2017-12-06T03:17:30

$V = 2 f t^{3}$ $V= 2ft^3$

Step 1: get equation for volume
$V = x {(3 - 2 x)}^{2}$ $V= x(3-2x)^2$

Step 2: Differentiate using chain rule and product rule
$V'=1*(3-2x)^2 + x(2)(3-2x)(-2)$
$V'=9-12x+4x^2-12x+8x^2$
$V'= 9-24x+12x^2$

Step 3: Find the critical numbers by find where V'=0 or V' DNE
$V' = 0= 9-24x+12x^2$
(note: V' DNE does not apply in this problem)

Step 4: factor to solve
$V' = 0= 9-24x+12x^2$
$0= 3(4x^2-6x+3)$
$0= (2x-3)(2x-1)$

$x=3/2, 1/2$

Step 5: rule out $x=3/2$ because it would cause the sides of the box to be negative. Plug $x=1/2$ into original volume equation.
$V(1/2)=2$