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Application of Derivatives question

2017 · Shift 0 · Q41

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Application of Derivatives
2017 · Shift 0 · Q41

Application of Derivatives question

2017 · Shift 0 · Q41

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is :

A
10
B
25
C
30
D
12.5

View written solutionFree

Correct answer: B

Form the perimeter constraint

Let the sector have radius $r$ and angle $\theta$ (in radians).

The wire is used for the boundary of the sector, which consists of:

two radii: $2r$
one arc: $r\theta$

So, $2r + r\theta = 20$ $r(\theta + 2)=20$ $r=\frac{20}{\theta+2}$

Write the area of the sector

Area of a sector is $A=\frac{1}{2}r^2\theta$

Substitute $r=\dfrac{20}{\theta+2}$ : $A(\theta)=\frac{1}{2}\left(\frac{20}{\theta+2}\right)^2\theta$ $A(\theta)=\frac{200\theta}{(\theta+2)^2}$

Maximize the area

Differentiate with respect to $\theta$ : $A(\theta)=200\theta(\theta+2)^{-2}$

Using product rule, $A'(\theta)=200\left[(\theta+2)^{-2}-2\theta(\theta+2)^{-3}\right]$

Take common factor $(\theta+2)^{-3}$ : $A'(\theta)=200(\theta+2)^{-3}\left[(\theta+2)-2\theta\right]$ $A'(\theta)=200(\theta+2)^{-3}(2-\theta)$

Set $A'(\theta)=0$ : $2-\theta=0$ $\theta=2$

Check maximum

Since for $\theta<2$ , $A'(\theta)>0$ , and for $\theta>2$ , $A'(\theta)<0$ , the area is maximum at $\theta=2$ .

Find the maximum area

At $\theta=2$ , $r=\frac{20}{2+2}=5$

Hence,

=\frac{1}{2}(5)^2(2)=25$$ So the maximum area is $$\boxed{25\text{ sq. m}}$$ --- 6. **Option check** - A: $10$ ❌ - B: $25$ ✅ - C: $30$ ❌ - D: $12.5$ ❌ Therefore, the correct option is **B**.

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