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Ncert Solutions Maths class 12th Maths Class 12th Application of Derivatives

91. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

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Vishal Baghel | Contributor-Level 10

4 months ago

Let ‘x’ cm be the length of side of the square to be cut off from the rectangular surface

Then, the volume v of the box is v = (45 - 2x) (24 - 2x) x

= 1080x - 138x² + 4x³

So, $\frac{d v}{d x} = 1080 - 276 x + 12 x^{2}$

$\frac{d^{2}}{d x^{2}} = - 276 + 24 x$

At, $\frac{d v}{d x} = 0 \to 1080 - 276 x + 12 x^{2} = 0$

$\Rightarrow$ x²- 23x + 90 = 0

$\Rightarrow$ x²- 5x - 18x + 90 = 0

$\Rightarrow$ x (x - 5) - 18 (x - 5) = 0

$\Rightarrow$ (x- 5) (x- 18) = 0

$\Rightarrow$ x = 5 and x = 18

At x = 18, breadth = 24 - 2 (18) = 24 - 36 = -12 which is not possible

At, $x = 5, \frac{d^{2} v}{d x^{2}} = - 276 + 24 (5) = - 276 + 120 = - 150 < 0$

Hence, x = 5 is the point of maximum

So, ‘5’ cm length of square seeds to be cut from each corner of the secthgle

Let ‘x’ cm be the length of side of the square to be cut off from the rectangular surfaceThen, the volume v of the box is v = (45 - 2x) (24 - 2x) x= 1080x - 138x2 + 4x3So,  <math><mrow><mfrac><mrow><mi>d</mi><mi>v</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mn>1</mn><mn>0</mn><mn>8</mn><mn>0</mn><mo>−</mo><mn>2</mn><mn>7</mn><mn>6</mn><mi>x</mi><mo>+</mo><mn>1</mn><mn>2</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></math><math><mrow><mfrac><mrow><msup><mrow><mi>d</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mi>d</mi><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mo>=</mo><mo>−</mo><mn>2</mn><mn>7</mn><mn>6</mn><mo>+</mo><mn>2</mn><mn>4</mn><mi>x</mi></mrow></math>At,  <math><mrow><mfrac><mrow><mi>d</mi><mi>v</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mn>0</mn><mo>→</mo><mn>1</mn><mn>0</mn><mn>8</mn><mn>0</mn><mo>−</mo><mn>2</mn><mn>7</mn><mn>6</mn><mi>x</mi><mo>+</mo><mn>1</mn><mn>2</mn><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>=</mo><mn>0</mn></mrow></math><math><mrow><mo>⇒</mo></mrow></math> x2- 23x + 90 = 0<math><mrow><mo>⇒</mo></mrow></math> x2- 5x - 18x + 90 = 0<math><mrow><mo>⇒</mo></mrow></math> x (x - 5) - 18 (x - 5) = 0<math><mrow><mo>⇒</mo></mrow></math> (x- 5) (x- 18) = 0<math><mrow><mo>⇒</mo></mrow></math> x = 5 and x = 18At x = 18, breadth = 24 - 2 (18) = 24 - 36 = -12 which is not possibleAt,  <math><mrow><mi>x</mi><mo>=</mo><mn>5</mn><mo>, </mo><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><mo>−</mo><mn>2</mn><mn>7</mn><mn>6</mn><mo>+</mo><mn>2</mn><mn>4</mn><mo stretchy="false"> (</mo><mn>5</mn><mo stretchy="false">)</mo><mo>=</mo><mo>−</mo><mn>2</mn><mn>7</mn><mn>6</mn><mo>+</mo><mn>1</mn><mn>2</mn><mn>0</mn><mo>=</mo><mo>−</mo><mn>1</mn><mn>5</mn><mn>0</mn><mo><</mo><mn>0</mn></mrow></math>Hence, x = 5 is the point of maximumSo, ‘5’ cm length of square seeds to be cut from each corner of the secthgle

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Let F₁(A,B,C) = $(A \land \sim B) \lor$ $[\sim C \land (A \lor B)] \lor \sim A a n d F_{2} (A, B) = (A \lor B) \lor (B \to \sim A)$ be two logical expressions. Then:

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From option let it be isosceles where AB = AC then

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Hence $Δ$ be equilateral having each side of length $\sqrt[]{3} r .$

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91. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?

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