JEE Main 2022 (27 June Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: The lengths of the sides of a triangle are $10 + x^{2}$ , $10 + x^{2}$ and $20 - 2 x^{2}$ . If for $x = k$ , the area of the triangle is maximum, then $3 k^{2}$ is equal to :

Q: Wait, why do we ignore the negative root when solving $x^2 = 10/3$?

Great observation! Remember, $x$ represents a physical dimension in our side lengths. If $x$ were negative, $x^2$ remains the same, so the geometry of the triangle is identical. We focus on the positive domain because the physical existence of the triangle is what matters, and the symmetry of the function handles the rest.

Select Answer:

Visualized Solution

Identify the Triangle Sides

Let the sides of the triangle be $a$ , $b$ , and $c$ .
$a = 10 + x^{2}$
$b = 10 + x^{2}$
$c = 20 - 2 x^{2}$
Note: This is an isosceles triangle since $a = b$ .

Calculate Semi-perimeter $s$

Semi-perimeter $s = \frac{a + b + c}{2}$
$s = \frac{( 10 + x ^{2} ) + ( 10 + x ^{2} ) + ( 20 - 2 x ^{2} )}{2}$

Simplify Semi-perimeter

$s = \frac{10 + x ^{2} + 10 + x ^{2} + 20 - 2 x ^{2}}{2}$
$s = \frac{40 + 2 x ^{2} - 2 x ^{2}}{2}$
$s = \frac{40}{2} = 20$

Apply Heron's Formula

Area $Δ = s (s - a) (s - b) (s - c)$
$s - a = 20 - (10 + x^{2}) = 10 - x^{2}$
$s - b = 20 - (10 + x^{2}) = 10 - x^{2}$
$s - c = 20 - (20 - 2 x^{2}) = 2 x^{2}$

Substitute into Area Formula

$Δ = 20 \cdot (10 - x^{2}) \cdot (10 - x^{2}) \cdot 2 x^{2}$
$Δ = 40 x^{2} (10 - x^{2})^{2}$

Simplify the Area Expression

$Δ = 40 x^{2} \cdot (10 - x^{2})^{2}$
$Δ = 210 ∣ x (10 - x^{2}) ∣$
Assuming $x > 0$ for maximum area, $Δ = 210 (10 x - x^{3})$

Define Function for Maximization

To maximize $Δ$ , we maximize the variable part $f (x) = 10 x - x^{3}$ .
Let $S = 10 x - x^{3}$ .
We need to find $x$ such that $\frac{d S}{d x} = 0$ .

Differentiate the Function

$\frac{d S}{d x} = \frac{d}{d x} (10 x - x^{3})$
$\frac{d S}{d x} = 10 - 3 x^{2}$

Solve for Critical Point

Set $\frac{d S}{d x} = 0 \Rightarrow 10 - 3 x^{2} = 0$
$3 x^{2} = 10$
$x^{2} = \frac{10}{3}$

Final Calculation for $3 k^{2}$

The area is maximum at $x = k$ , so $k^{2} = \frac{10}{3}$ .
We need to find the value of $3 k^{2}$ .
$3 k^{2} = 3 \cdot (\frac{10}{3}) = 10$
Final Answer: 10

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The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

We are given a triangle with side lengths defined by

a = 10 + x^{2}

b = 10 + x^{2}

, and

c = 20 - 2 x^{2}

Observing the structure, we identify this as an isosceles triangle. This symmetry is our first hint that the algebraic complexity will simplify significantly.

Calculating the Semi-Perimeter

To find the area, we invoke Heron's Formula:

Δ = s (s - a) (s - b) (s - c)

First, we calculate the semi-perimeter

s

s = \frac{( 10 + x ^{2} ) + ( 10 + x ^{2} ) + ( 20 - 2 x ^{2} )}{2}

Notice that the

x^{2}

terms cancel out perfectly. This yields a constant value for the semi-perimeter:

s = \frac{40}{2} = 20

The Master Equation

With

s = 20

, we calculate the individual components of Heron's Formula:

s - a = 20 - (10 + x^{2}) = 10 - x^{2}

s - b = 20 - (10 + x^{2}) = 10 - x^{2}

s - c = 20 - (20 - 2 x^{2}) = 2 x^{2}

Substituting these into the area formula, we obtain:

Δ = 20 \cdot (10 - x^{2}) \cdot (10 - x^{2}) \cdot 2 x^{2}

Simplifying the expression, we get:

Δ = 40 \cdot x^{2} \cdot (10 - x^{2})^{2} = 210 \cdot x (10 - x^{2})

Final Calculation

To maximize the area, we focus on maximizing the function

f (x) = 10 x - x^{3}

We differentiate

f (x)

with respect to

x

f^{'} (x) = 10 - 3 x^{2}

Setting the derivative to zero to find the critical point:

10 - 3 x^{2} = 0 \Rightarrow 3 x^{2} = 10

The problem asks for the value of

3 k^{2}

(where

k

represents the value of

x

at the maximum). Thus, the final result is:

$3 k^{2} = 10$

Common Traps & FAQs

Why can't we just use the base-height formula? It seems easier.

You absolutely can! But Heron's formula is a beautiful, symmetric tool that handles the algebra of all three sides simultaneously. In competitive exams, choosing the path of least resistance—where the algebra simplifies itself—is a skill. Heron's formula here leads to a clean cancellation that makes the calculus trivial.

Wait, why do we ignore the negative root when solving

x^{2} = 10/3

Great observation! Remember,

x

represents a physical dimension in our side lengths. If

x

were negative,

x^{2}

remains the same, so the geometry of the triangle is identical. We focus on the positive domain because the physical existence of the triangle is what matters, and the symmetry of the function handles the rest.

Do I need to check the second derivative to confirm it's a maximum?

In a formal proof, yes! Always check the second derivative to ensure concavity. Here,

f^{''} (x) = - 6 x

. For

x = 10/3

, the second derivative is negative, confirming a local maximum. It's a great habit to build for JEE Advanced!

Animated Solution for Mathematics - Differentiation: The lengths of the sides of a triangle are $10 + x^{2}$ , $10 + x^{2}$ and $20 - 2 x^{2}$ . If for $x = k$ , the area of the triangle is maximum, then $3 k^{2}$ is equal to :

Select Answer:

Visualized Solution

Identify the Triangle Sides

Calculate Semi-perimeter $s$

Simplify Semi-perimeter

Apply Heron's Formula

Substitute into Area Formula

Simplify the Area Expression

Define Function for Maximization

Differentiate the Function

Solve for Critical Point

Final Calculation for $3 k^{2}$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Calculating the Semi-Perimeter

The Master Equation

Final Calculation

Similar Questions

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Let a rectangle $A B C D$ of sides 2 and 4 be inscribed in another rectangle $P QR S$ such that the vertices of the rectangle $A B C D$ lie on the sides of the rectangle $P QR S$ . Let $a$ and $b$ be the sides of the rectangle $P QR S$ when its area is maximum. Then $(a + b)^{2}$ is equal to :

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Select Answer:

Visualized Solution

Identify the Triangle Sides

Calculate Semi-perimeter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s

Simplify Semi-perimeter

Apply Heron's Formula

Substitute into Area Formula

Simplify the Area Expression

Define Function for Maximization

Differentiate the Function

Solve for Critical Point

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3k2

The Sigma Insight: Maxima and Minima

Analyzing the Setup

Calculating the Semi-Perimeter

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a rectangle is inscribed in an equilateral triangle of side length 22​ as shown in the figure, then the square of the largest area of such a rectangle is ___

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The maximum area of a triangle whose one vertex is at (0,0) and the other two vertices lie on the curve y=−2x2+54 at points (x,y) and (−x,y) where y>0 is :

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A point P is given on the circumference of a circle of radius r. Chord QR is parallel to the tangent at P. Determine the maximum possible area of the triangle PQR.

A wire of length 22 m 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 :

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:

Calculate Semi-perimeter $s$

Final Calculation for $3 k^{2}$

If a rectangle is inscribed in an equilateral triangle of side length $22$ as shown in the figure, then the square of the largest area of such a rectangle is ___

The maximum area of a triangle whose one vertex is at $(0, 0)$ and the other two vertices lie on the curve $y = - 2 x^{2} + 54$ at points $(x, y)$ and $(- x, y)$ where $y > 0$ is :

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