Question

Consider the following problem: A box with an open top 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.

Answer 1

The largest volume that such can have is 2 cubic feet.

Explanation:

Given : A box with an open top 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.

To find : The largest volume that such a box can have ?

Solution :

Let the side of the corner be 'a' ft and the side of the base be 'b' ft.

The height of the box is 'a' ft.

The volume of the box is
`V=L* B* H`

`V=ab^2`

The total length of the board is 3 ft.

So, b=3-2a ft.

Substitute in volume,

`V=a(3-2a)(3-2a)`

`V=a(9-12a+4a^2)`

`V(a)=9a-12a^2+4a^3`

Differentiate w.r.t. a,

`V'(a)=9-24a+12a^2`

For critical point put V'(a)=0,

`12a^2-24a+9=0`

`4a^2-8a+3=0`

`4a^2-2a-6a+3=0`

`(2a-3)(2a-1)=0`

`a=(3)/(2),(1)/(2)`

Differentiate again w.r.t to a,

`V''(a)=-24+24a`

`V''((3)/(2))=-24+24((3)/(2))=12>0` volume is minimum.

`V''((1)/(2))=-24+24((1)/(2))=-12<0` volume is maximum.

The volume is maximum at
`a=(1)/(2)`

So, the maximum volume is given by

`V=a(3-2a)^2`

`V=(1)/(2)(3-2((1)/(2)))^2`

`V=(1)/(2)(3-1)^2`

`V=(1)/(2)(2)^2`

`V=2\ ft^3`

Therefore, the largest volume that such can have is 2 cubic feet.

Answer 2

Thus, maximum value of box is 2 square feet at
`x = (1)/(2)`.

Explanation:

A square piece of cardboard of side 3 feet is cut in such that a square is cut from each corner. Let x be the side of this square cut. When it was folded to make the box the height of box becomes x, length becomes (3-2x) and the width becomes (3-2x).

Volume is given by

V =
`V = Length* Width* Height\\V = (3 - 2x)^2x = x(4x^2-12x+9) = 4x^3-12x^2+9x\\So,\\V(x) = 4x^3-12x^2+9x`

First, we differentiate V(x) with respect to x, to get,

`(d(V(x)))/(dx) = (d(4x^3-12x^2+9x))/(dx) = 12x^2 - 24x +9`

Equating the first derivative to zero, we get,

`(d(V(x)))/(dx) = 0\\\\12x^2 - 24x +9 = 0`

Solving, with the help of quadratic formula, we get,

`x =(3)/(2), (1)/(2)`

Again differentiation V(x), with resopect to x, we get,

`(d^2(V(x)))/(dx^2) = 24x - 24`

At x =
`(1)/(2)`,

`(d^2(V(x)))/(dx^2) < 0`

Thus, maxima occurs at x =
`(1)/(2)` for V(x).

Thus, largest volume the box can have occurs when
`x = (1)/(2)`.

Maximum value of volume

V(
`(1)/(2)`) =
`(3-1)^2* (1)/(2) = 2\text{ square feet}`

Thus, maximum value of box is 2 square feet at
`x = (1)/(2)`.