The maximum area of a triangle whose one vert | vedclass

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(D) The vertices of the triangle are $(0,0)$,$(x, y)$,and $(-x, y)$.The area of the triangle is given by the determinant formula:$	ext{Area} = \frac{

Vedclass · Accepted Answer

$108$

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(D) The vertices of the triangle are $(0,0)$,$(x, y)$,and $(-x, y)$.The area of the triangle is given by the determinant formula:$	ext{Area} = \frac{1}{2} |0(y - y) + x(y - 0) + (-x)(0 - y)|$$	ext{Area} = \frac{1}{2} |xy + xy| = |xy|$Given the curve $y = -2x^2 + 54$,we substitute $y$ into the area expression:$A(x) = |x(-2x^2 + 54)| = |-2x^3 + 54x|$Since $y > 0$,we have $-2x^2 + 54 > 0$,which implies $x^2 For $x > 0$,$A(x) = -2x^3 + 54x$.To find the maximum area,we differentiate $A(x)$ with respect to $x$:$\frac{dA}{dx} = -6x^2 + 54$Setting $\frac{dA}{dx} = 0$:$-6x^2 + 54 = 0 \Rightarrow x^2 = 9 \Rightarrow x = 3$ (since $x > 0$).Now,calculate the maximum area at $x = 3$:$A(3) = |3(-2(3)^2 + 54)| = |3(-18 + 54)| = |3(36)| = 108$.Thus,the maximum area of the triangle is $108$.

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

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