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Application of Derivatives question

2024 · 30 Jan · Shift 1 · Q35

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Application of Derivatives
2024 · 30 Jan · Shift 1 · Q35

Application of Derivatives question

2024 · 30 Jan · Shift 1 · Q35

JEE MainMathematicsApplication of DerivativesMCQ+4 / −1

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\gt 0

, is :

A
108
B
122
C
88
D
92

View written solutionFree

Correct answer: A

Coordinates of the triangle

The triangle has vertices:

$O=(0,0)$
$P=(x,y)$
$Q=(-x,y)$

Since $P$ and $Q$ lie on the curve $y=-2x^2+54,$ we have

\qquad y>0.$$ 2. **Area of the triangle** The points $P$ and $Q$ have the same $y$-coordinate, so $PQ$ is a horizontal line segment. - Base $PQ = x-(-x)=2x$ - Height from $(0,0)$ to the line $y=$ constant is $y$ Hence the area is $$A=\frac{1}{2}(2x)(y)=xy.$$ Now substitute $y=-2x^2+54$: $$A(x)=x(-2x^2+54)=-2x^3+54x.$$ Since area must be positive, we take $x>0$. 3. **Maximize the area** Differentiate: $$A'(x)=\frac{d}{dx}(-2x^3+54x)=-6x^2+54.$$ Set $A'(x)=0$: $$-6x^2+54=0$$ $$6x^2=54$$ $$x^2=9$$ $$x=3$$ (using $x>0$). 4. **Check that it is a maximum** Second derivative: $$A''(x)=-12x.$$ At $x=3$, $$A''(3)=-36<0,$$ so $A$ is maximum at $x=3$. 5. **Compute the maximum area** First find $y$: $$y=-2(3)^2+54=-18+54=36.$$ Therefore, $$A_{\max}=xy=(3)(36)=108.$$ 6. **Compare with options** The maximum area is $$\boxed{108}.$$ So the correct option is **A**.

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