You have a open box that is made from a 16 in. x30 in. piece of cardboard. When you cut out the squares of equal size from the 4 corners and bending it. What size should the squares be to get this box to work with the largest volume?

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Answer 1 · 2018-03-04T12:30:50

$3 \frac{1}{3}$ $3 1/3$ inches to be cut from $4$ $4$ corners and bend to get
box for maximum volume of $725.93$ $725.93$ cubic inches.

Card board size is $L = 30 and W = 16$ $L=30 and W=16$ inches

Let $x$ $x$ in square is cut from $4$ $4$ corners and bended into

a box whos size is now $L = 30 - 2 x, W = 16 - 2 x and h = x$ $L=30-2x , W=16-2x and h=x$

inches. Volume of the box is $V = (30 - 2 x) (16 - 2 x) x$ $V=(30-2x)(16-2x)x$ cubic

inches. $V = (4 x^{2} - 92 x + 480) x = 4 x^{3} - 92 x^{2} + 480 x$ $V=(4x^2-92x+480)x = 4x^3-92x^2+480x$ .

For maximum value $\frac{d V}{d x} = 0$ $(dV)/dx=0$

$\frac{d V}{d x} = 12 x^{2} - 184 x + 480 = 12 (x^{2} - \frac{46}{3} x + 40)$ $(dV)/dx=12x^2-184x+480=12(x^2-46/3x+40)$

$12 (x^{2} - 12 x - \frac{10}{3} x + 40) = 12 (x (x - 12) - \frac{10}{3} (x - 12))$ $12(x^2-12x-10/3x+40)= 12(x(x-12)-10/3(x-12))$

or $12(x-12)(x-10/3)=0 :.$ Critical points are

$x=12 ,x=10/3; x !=12$ , as $24$ inches cannot be removed from

$16$ inches width. So $x= 10/3 or 3 1/3$ inches to be cut.

Slope test may be examined at $(x=3 and x=4)$ to show

volume is maximum. $(dV)/dx=12(x-12)(x-10/3)$

$(dV)/dx(3)= (+) and (dV)/dx(4)= (-)$ . Slope at critical point

is from positive to negative , so the volume is maximum.

The maximum volume is $V=(30-20/3)(16-20/3)10/3$ or

$V=(30-20/3)(16-20/3)10/3 ~~725.93$ cubic inches. [Ans]