Question

You are creating an open top box with a piece of cardboard that is 16 x 30“. What size of square should be cut out of each corner to create a box with the largest volume?

Answer 1

Answer:

$\frac{10}{3} \ inches$ of square should be cut out of each corner to create a box with the largest volume.

Step-by-step explanation:

Given: Dimension of cardboard= 16 x 30“.

As per the dimension given, we know Lenght is 30 inches and width is 16 inches. Also the cardboard has 4 corners which should be cut out.

Lets assume the cut out size of each corner be "x".

∴ Size of cardboard after 4 corner will be cut out is:

Length (l)= $30-2x$

Width (w)= $16-2x$

Height (h)= $x$

Now, finding the volume of box after 4 corner been cut out.

Formula; Volume (v)= $l\times w\times h$

Volume(v)= $(30-2x)\times (16-2x)\times x$

Using distributive property of multiplication

⇒ Volume(v)= $4x^{3} -92x^{2} +480x$

Next using differentiative method to find box largest volume, we will have $\frac{dv}{dx}= 0$

$\frac{d (4x^{3} -92x^{2} +480x)}{dx} = \frac{dv}{dx}$

Differentiating the value

⇒$\frac{dv}{dx} = 12x^{2} -184x+480$

taking out 12 as common in the equation and subtituting the value.

⇒ $0= 12(x^{2} -\frac{46x}{3} +40)$

solving quadratic equation inside the parenthesis.

⇒$12(x^{2} -12x-\frac{10x}{x} +40)$=0

Dividing 12 on both side

⇒$[x(x-12)-\frac{10}{3} (x-12)]$= 0

We can again take common as (x-12).

⇒ $x(x-12)[x-\frac{10}{3} ]$=0

∴$(x-\frac{10}{3} ) (x-12)= 0$

We have two value for x, which is $12 and \frac{10}{3}$

12 is invalid as, w= $(16-2x)= 16-2\times 12$

∴ 24 inches can not be cut out of 16 inches width.

Hence, the cut out size from cardboard is $\frac{10}{3}\ inches$

Now, subtituting the value of x to find volume of the box.

Volume(v)= $(30-2x)\times (16-2x)\times x$

⇒ Volume(v)= $(30-2\times \frac{10}{3} )\times (16-2\times \frac{10}{3})\times \frac{10}{3}$

⇒ Volume(v)= $(30-\frac{20}{3} ) (16-\frac{20}{3}) (\frac{10}{3} )$

∴  Volume(v)= 725.93 inches³