Question

Six identical squares are cut from the corners and edges of an 80 cm by 50 cm cardboard rectangle. the remaining piece is folded into a closed box with 3 flaps. what is the largest possible volume of such a box?

Answer 1

Solution:

The Given Rectangle having dimensions

Length = 80 cm

Breadth = 50 cm

Let the Six squares which has been cut from this rectangle have side of length a cm.

Area of each square = (Side)²= a²

Area of 6 Identical Squares = 6 × a²= 6 a²

If four squares are cut from four corners and two along Length,

then , Length of Box = (80 - 3 a)cm, Breadth of Box = (50 - 2 a)cm, Height = a cm

Volume of Box =V = Length × Breadth × Height

    V = (80 - 3 a)× (50 - 2 a)× a

     V   = 4000 a - 310 a² + 6 a³

For Maximum Volume

V'= 0 , where V' = Derivative of V with respect to a.

V'= 4000 - 620 a + 18 a²

V' =0

18 a² - 620 a + 4000= 0

9 a² - 310 a + 2000=0

using determinant method

$a = \frac{310\pm\sqrt{96100-72000}}{18}=\frac{310\pm155}{18}=\frac{310-155}{18}=8.6$ cm

V"=1 8 a - 310 = -ve

which shows , when a = 8.6 cm , volume is maximum.

So, V = 4000×8.6 - 310×(8.6)²+6×(8.6)³=15288.736 cm³

OR

If four squares are cut from four corners and two along Breadth,

then , Length of Box = (80 - 2 a)cm, Breadth of Box = (50 -  3 a)cm, Height = a cm

Volume of Box =V = Length × Breadth × Height

    V = (80 - 2 a)× (50 - 3 a)× a

     V   = 4000 a - 340 a² + 6 a³

For Maximum Volume

V'= 0 , where V' = Derivative of V with respect to a.

V'= 4000 - 680 a + 18 a²

V' =0

18 a² - 680 a + 4000= 0

9 a² - 340 a + 2000=0

Using Determinant method

$a = \frac{340\pm\sqrt{115600-72000}}{18}=\frac{340\pm209}{18}=\frac{131}{18}=7.6$ cm

V"=18 a -340= -ve value when a = 7.6 cm, shows volume is maximum when a = 7.6 cm

V= 4000×7.6 -340 × (7.6)² +6× (7.6)³=13395.456 cubic cm

Answer 2

Check the picture.

let the length of a side of each of the squares removed be x.

The box formed will have dimensions: 80-2x, 50-2x, x(the height)

So the volume can be expressed as a function of x as follows:

f(x)=(80-2x)(50-2x)x=$[4000-160x-100x+4 x^{2} ]x=(4 x^{2}-260x+4000)x$

so $f(x)=4 x^{3}-260x^{2}+4000x$

the solutions of f'(x)=0 gives the inflection points, so the candidates for maxima points,

$f'(x)=12x^{2}-520 x +4000=0$

solving the quadratic equation, either by a calculator, graphing software, or by other algebraic methods as the discriminant formula, we find the solutions

x=10 and x=33.333

plug in f(x) these values to see which greater:

$f(10)=(80-20)(50-20)10=60*30*10=18000$ cm cubed

$f(33.333)=(80-66.666)(50-66.666)33.333=$ which is negative because (50-66.666)<0



Answer: 18000 cm cubed