Let A be the region enclosed by the parabola | vedclass

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(A) Let the rectangle have vertices at $(x, b)$,$(24, b)$,$(24, -b)$,and $(x, -b)$.Since the vertex $(x, b)$ lies on the parabola $y^2=2x$,we have $b^

Vedclass · Accepted Answer

$128$

Vedclass · Answer

(A) Let the rectangle have vertices at $(x, b)$,$(24, b)$,$(24, -b)$,and $(x, -b)$.Since the vertex $(x, b)$ lies on the parabola $y^2=2x$,we have $b^2=2x$,which implies $x = \frac{b^2}{2}$.The width of the rectangle is $(24 - x) = (24 - \frac{b^2}{2})$ and the height is $2b$.The area $A$ of the rectangle is given by $A = 2b(24 - \frac{b^2}{2}) = 48b - b^3$.To find the maximum area,we differentiate $A$ with respect to $b$:$\frac{dA}{db} = 48 - 3b^2$.Setting $\frac{dA}{db} = 0$,we get $3b^2 = 48$,so $b^2 = 16$,which gives $b = 4$ (since $b > 0$).The maximum area is $A = 48(4) - (4)^3 = 192 - 64 = 128$.

Let $A$ be the region enclosed by the parabola $y^2=2x$ and the line $x=24$. Then the maximum area of the rectangle inscribed in the region $A$ is ..................

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