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

113. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

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Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

Let ‘x’ metre be the radius of the semi-circular opening mounded on the length ‘2x’ side of rectangle. Then, let ‘y’ be the breadth of the rectangle.

Then, perimeter of the window = 10m

$\Rightarrow \frac{2 π x}{2} + 2 x + y + y = 10$

x + 2x + 2y + = 10.

$\Rightarrow y = \frac{10 - (π + 2) π}{2 .}$

Let the area of the window be A.

Then, A = $\frac{1}{2} (π x^{2}) + 2 x y .$

$= \frac{π x^{2}}{2} + 2 x \cdot [\frac{10 - (x + 2) x}{2}] .$

$= \frac{π x^{2} + 20 x - 2 π x^{2} - 4 x^{2}}{2}$

= $\frac{1}{2}$ [-πx²- 4x² + 20x].

So, $\frac{d}{d x} = \frac{1}{2}$ [ -2πx - 8x + 20]

And $\frac{d \overset{2}{}}{d x^{2}} = \frac{1}{2}$ [ -2π - 8] = -π -4 = -( π+ 4)

At $\frac{d}{d x} = 0 .$

$\frac{1}{2}$ [ -2πx - 8x + 20] = 0

2x + 8x = 20

x = $\frac{20}{2 π + 8}$ = $\frac{10}{π + 4} .$

At x = $\frac{10}{π + 4}, \frac{d^{2A}}{d x^{2}}$ = -( π+ 4) < 0

Øx = $\frac{10}{x + 4}$ is a point of minima.

And y = $\frac{10 - (x + 2) \times (\frac{10}{x + 4})}{2}$

$= \frac{10 (π + 4) - (π + 2) 10}{2 (π + 4)}$

$= \frac{10 π + 40 - 10 π - 20}{2 (x + 4)} = \frac{10}{π + 4 .}$

Ø Dimensions of the window are

length = 2x = $\frac{20}{π + 4} m$

breadth = y $\frac{10}{π + 4} m .$

radius = y = $\frac{10}{π + 4} m .$

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y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
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2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
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The lengths of the sides of a triangle are 10 + x2, 10 + x2 and 20 – 2x2. If for x = k, the area of the triangle is maximum, then 3k2 is equal to:

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$C D = \sqrt[]{{(10 + x^{2})}^{2} - {(10 - x^{2})}^{2}} = 2 \sqrt[]{10} | x |$

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$= \frac{1}{2} \times C D \times A B = \frac{1}{2} \times 2 \sqrt[]{10} | x | (20 - 2 x^{2})$

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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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By truth table

So F₁ (A, B, C) is not a tautology

Now again by truth table

So F₂ (A, B) be a tautology.

The triangle of maximum area that can be inscribed in a given circle of radius ‘r’ is:

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From option let it be isosceles where AB = AC then

$x = \sqrt[]{r^{2} - {(h - r)}^{2}}$

= $\sqrt[]{r^{2} - h^{2} - r^{2} + 2 r h}$

$x = \sqrt[]{2 h r - h^{2}} . . . . . . . . (i)$

Now ar $(Δ A B C) = Δ = \frac{1}{2} B C \times A L$

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then $x = \sqrt[]{2 \times \frac{3 r}{2} \times r - \frac{9 r^{2}}{4}} = \frac{\sqrt[]{3}}{2} r f r o m (i)$

$\Rightarrow B C = \sqrt[]{3} r$

So $A B = \sqrt[]{h^{2} + x^{2}} = \sqrt[]{\frac{9 r^{2}}{4} + \frac{3 r^{2}}{4}} = \sqrt[]{3} r$ .

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

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113. A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

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