1. **Problem Statement:** We have a point $M$ on the parabola given by the equation $$y = \frac{1}{2}x^2 - 2x + 3.$$ We draw perpendiculars from $M$ to the $x$-axis and $y$-axis, forming a rectangle in the first quadrant. We want to find the minimum perimeter of this rectangle. 2. **Identify the rectangle sides:** The coordinates of $M$ are $(x,y)$ with $x > 0$ and $y > 0$ since the rectangle is in the first quadrant. The rectangle has vertices at $(0,0)$, $(x,0)$, $(x,y)$, and $(0,y)$. 3. **Perimeter expression:** The perimeter $P$ of the rectangle is $$P = 2(x + y).$$ 4. **Substitute $y$ using the parabola:** Since $y = \frac{1}{2}x^2 - 2x + 3$, we get $$P = 2\left(x + \frac{1}{2}x^2 - 2x + 3\right) = 2\left(-x + \frac{1}{2}x^2 + 3\right) = x^2 - 2x + 6.$$ 5. **Domain considerations:** For the rectangle to be in the first quadrant, we require $x > 0$ and $y > 0$. Let's check where $y > 0$: $$\frac{1}{2}x^2 - 2x + 3 > 0.$$ Since this is a parabola opening upwards, its minimum is at $x = 2$, $$y(2) = \frac{1}{2}(2)^2 - 2(2) + 3 = 2 - 4 + 3 = 1 > 0,$$ so $y$ is positive for all relevant $x > 0$. 6. **Find minimum perimeter:** Minimize $$P = x^{2} - 2x + 6$$ for $x > 0$. 7. **Derivative and critical points:** $$\frac{dP}{dx} = 2x - 2.$$ Set to zero to find critical points: $$2x - 2 = 0 \implies x = 1.$$ 8. **Second derivative test:** $$\frac{d^2P}{dx^2} = 2 > 0,$$ so $x=1$ is a minimum. 9. **Minimum perimeter value:** $$P_{min} = 1^2 - 2(1) + 6 = 1 - 2 + 6 = 5.$$ **Answer:** The minimum perimeter of the rectangle formed in the first quadrant is $\boxed{5}$.