Question

A rectangular box with an open top is to be made from a 14 14-in.-by- 30 30-in. piece of cardboard by removing small squares of equal size from the corners and folding up the remaining flaps. What should be the size of the squares cut from the corners so that the box will have the largest possible volume

Answer 1

Answer:

Therefore 2.54 in be the size of the square cut from the corner of the cardboard.

Step-by-step explanation:

Given that, a rectangle box is to be made from 14 in by 30 in piece of cardboard with an open top.

The side of the square be x(say) in length.

The length of the rectangular box is =(14 - 2x) in

The width of the rectangular box is = (30-2x) in

The height of the rectangular box is = x in.

The volume of the rectangular box is =(14-2x)(30-2x)x $in^3$

∴V=(14-2x)(30-2x)x [ where V is in cubic inches]

    =420x- 88x²+4x³

Differentiating with respect to x

V' = 420 -196x+12x²

Again differentiating with respect to x

V''= -196+24x

We set V' = 0 , to find the dimension of the for which the volume the box will be maximum.

∴420 -196x+12x²=0

⇒4(3x²-49x+105)=0

⇒3x²-49x+105=0

Apply the quadratic formula to find the value of x. The formula is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, here a= 3, b= -49 , c=105

$\Rightarrow x=\frac{-(-49)\pm\sqrt{(-49)^2-4.3.105}}{2.3}$

⇒x=13.80, 2.54

For x= 13.80 , the length of the box will be negative.

∴x=2.54

$\therefore V''|_{x=2.54}= -196+(24\times 2.54)=-135.04<0$

Therefore when x=2.54 , the volume of the box will be maximum.

Therefore 2.54 in be the size of the square cut from the corner of the cardboard.