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

A wire of length 20m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is:

Option 1 -

$\frac{5}{3 + \sqrt[]{3}}$

Option 2 -

$\frac{10}{3 + 2 \sqrt[]{3}}$

Option 3 -

$\frac{10}{2 + 3 \sqrt[]{3}}$

Option 4 -

$\frac{5}{2 + \sqrt[]{3}}$

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Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

Let square is made with piece of length x metre & hexagon with piece of length y metre

x + y = 20 .(i)

$a = \frac{x}{4} & b = \frac{y}{6}$

Now let A = area of square + area of hexagon

$A = \frac{x^{2}}{16} + 6 \times \frac{\sqrt[]{3}}{4} \times \frac{y^{2}}{36} = \frac{x^{2}}{16} + \frac{\sqrt[]{3} y^{2}}{24} = \frac{x^{2}}{16} + \frac{\sqrt[]{3}}{24} {(20 - x)}^{2} f r o m (i)$ ,

for minimum area

$x = \frac{80 \sqrt[]{3}}{6 + 4 \sqrt[]{3}} = \frac{80 \sqrt[]{3}}{2 \sqrt[]{3} (\sqrt[]{3} + 2)} = \frac{40}{2 + \sqrt[]{3}}$

$x = 40 (2 - \sqrt[]{3})$

=> side of hexagon = $\frac{y}{6} = \frac{20 \sqrt[]{3} (2 - \sqrt[]{3})}{6} = \frac{20 \sqrt[]{3}}{6 (2 + \sqrt[]{3})} = \frac{10 \sqrt[]{3}}{3 (2 + \sqrt[]{3})} = \frac{10}{2 \sqrt[]{3} + 3}$

Let square is made with piece of length x metre & hexagon with piece of length y metrex + y = 20 .(i) <math> <mrow> <mi>a</mi> <mo>=</mo> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>&</mo> <mi>b</mi> <mo>=</mo> <mfrac> <mrow> <mi>y</mi> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> </mrow> </math>       Now let A = area of square + area of hexagon <math> <mrow> <mi>A</mi> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>6</mn> </mrow> </mfrac> <mo>+</mo> <mn>6</mn> <mo>×</mo> <mfrac> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mo>×</mo> <mfrac> <mrow> <msup> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> <mn>6</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>6</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <msup> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>2</mn> <mn>4</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>6</mn> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mrow> <mn>2</mn> <mn>4</mn> </mrow> </mfrac> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mn>0</mn> <mo>−</mo> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mi>f</mi> <mi>r</mi> <mi>o</mi> <mi>m</mi> <mrow> <mo>(</mo> <mrow> <mi>i</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> ,for minimum area <math> <mrow> <mi>x</mi> <mo>=</mo> <mfrac> <mrow> <mn>8</mn> <mn>0</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mrow> <mn>6</mn> <mo>+</mo> <mn>4</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>8</mn> <mn>0</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mrow> <mn>2</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mrow> <mo>(</mo> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>4</mn> <mn>0</mn> </mrow> <mrow> <mn>2</mn> <mo>+</mo> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> </math>  <math> <mrow> <mi>x</mi> <mo>=</mo> <mn>4</mn> <mn>0</mn> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mo>−</mo> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mo>)</mo> </mrow> </mrow> </math> => side of hexagon = <math> <mrow> <mfrac> <mrow> <mi>y</mi> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mn>0</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mo>−</mo> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>6</mn> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mn>0</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mrow> <mn>6</mn> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mo>+</mo> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mrow> <mn>3</mn> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mo>+</mo> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> </mrow> <mrow> <mn>2</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mn>3</mn> </mrow> </mfrac> </mrow> </math>

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A wire of length 20m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is:

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