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90. A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.

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Vishal Baghel | Contributor-Level 10

4 months ago

Side of the tin square piece = 18 cm

Let x cm be the thought of the square to be cut from each corner.

The volume v of the box after cutting

$\Rightarrow$ v = length × breadth × height

= (18 - 2x) (18 - 2x) x x

= (18 - 2x)²x

= (324 + 4x²- 72x) x

= 4x³- 72x² + 324x

So, $\frac{d}{d x} = 12 x^{2} - 144 x + 324$

$\frac{d^{2} v}{d x^{2}} = 24 x - 144$

As $\frac{d}{d x} = 0 \to 12 x^{2} - 144 x + 324 = 0$

$\Rightarrow$ x²- 12x + 27 = 0

$\Rightarrow$ x²- 9x- 3x + 27 = 0

$\Rightarrow$ x (x - 9) - 3 (x - 9) = 0

$\Rightarrow$ (x - 9) (x - 3) = 0

$\Rightarrow$ x = 9 and x = 3

At x = 0, length of box = 18 - 2π9 = 18 - 18 = 0

Which is not possible

And at x = 3, $\frac{d^{2} y}{d x^{2}}$ = 24 (3) - 144 = -72 < 0

∴x = 3 is a point of maximum

Hence, ‘3’ cm (square) is to be cut from each side of the square

So that volume of box is maximum

Side of the tin square piece = 18 cmLet x cm be the thought of the square to be cut from each corner.The volume v of the box after cutting<math><mrow><mo>⇒</mo></mrow></math> v = length × breadth × height= (18 - 2x) (18 - 2x)  x x= (18 - 2x)2x= (324 + 4x2- 72x) x= 4x3- 72x2 + 324xSo,  <math><mrow><mfrac><mrow><mi>d</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mn>1</mn><mn>2</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>1</mn><mn>4</mn><mn>4</mn><mi>x</mi><mo>+</mo><mn>3</mn><mn>2</mn><mn>4</mn></mrow></math><math><mrow><mfrac><mrow><msup><mrow><mi>d</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>v</mi></mrow><mrow><mi>d</mi><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mo>=</mo><mn>2</mn><mn>4</mn><mi>x</mi><mo>−</mo><mn>1</mn><mn>4</mn><mn>4</mn></mrow></math>As <math><mrow><mfrac><mrow><mi>d</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mn>0</mn><mo>→</mo><mn>1</mn><mn>2</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>1</mn><mn>4</mn><mn>4</mn><mi>x</mi><mo>+</mo><mn>3</mn><mn>2</mn><mn>4</mn><mo>=</mo><mn>0</mn></mrow></math><math><mrow><mo>⇒</mo></mrow></math> x2- 12x + 27 = 0<math><mrow><mo>⇒</mo></mrow></math>x2- 9x- 3x + 27 = 0<math><mrow><mo>⇒</mo></mrow></math> x (x - 9) - 3 (x - 9) = 0<math><mrow><mo>⇒</mo></mrow></math> (x - 9) (x - 3) = 0<math><mrow><mo>⇒</mo></mrow></math> x = 9 and x = 3At x = 0, length of box = 18 - 2π9 = 18 - 18 = 0Which is not possibleAnd at x = 3,  <math><mrow><mfrac><mrow><msup><mrow><mi>d</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>y</mi></mrow><mrow><mi>d</mi><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac></mrow></math> = 24 (3) - 144 = -72 < 0∴x = 3 is a point of maximumHence, ‘3’ cm (square) is to be cut from each side of the squareSo that volume of box is maximum

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90. A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.

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