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Maths Class 12th Application of Derivatives Maths NCERT Exemplar Solutions Class 12th Chapter Six

A wire of length 22m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is:

Option 1 -

$\frac{22}{9 + 4 \sqrt[]{3}}$

Option 2 -

$\frac{66}{9 + 4 \sqrt[]{3}}$

Option 3 -

$\frac{22}{4 + 9 \sqrt[]{3}}$

Option 4 -

$\frac{66}{4 + 9 \sqrt[]{3}}$

3 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

Let perimeter of $Δ$ is x and that of square is 22 – x

now area $= \frac{\sqrt[]{3}}{4} {(\frac{x}{3})}^{2} + {(\frac{22 - x}{4})}^{2}$

for maximum or minimum, $\frac{d A}{d x} = 0$

x $= \frac{22 \sqrt[]{3}}{\sqrt[]{3} + 4}$

now side of a $Δ = \frac{x}{3}$

$= \frac{22 \sqrt[]{3}}{3 (\sqrt[]{3} + 4)}$

$= 66$ $9$ $+$ $4$

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

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A wire of length 22m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is:

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