Question

A figure is bounded by the curves \(y=x^{2}+1\), \(y=0\), \(x=0\) and \(x=1\). The point at which a tangent should be drawn to the curve \(y=x^{2}+1\) for it to cut off a trapezium of the greatest area from the figure is:

• A. $(1,2)$
• B. $(-1,2)$
• C. $\left(\frac{1}{2},\frac{5}{4}\right)$
• D. $(0,1)$

Hint:

The area function $A(t) = 1 + t - t^2$ forms an downward-opening parabola. Its absolute vertex turning point occurs exactly at the midpoint of the horizontal range interval, which lets you identify $t = 1/2$ by inspection!

Answer 1

The Correct Option is C

Concept: To maximize the area of a trapezium cut off by a tangent line to a curve over an interval $[0, 1]$, we determine the equation of the tangent line at an arbitrary parameter point $x = t$. The area under a linear line $Y = mX + c$ over the interval $[0, 1]$ is evaluated using integration or by calculating the geometric mean height: $\text{Area} = \frac{Y(0) + Y(1)}{2} \cdot 1$. Step 1: Determine the equation of the tangent line at a point.
Let the tangent line be drawn to the parabola $y = x^2 + 1$ at a candidate point $P(t, t^2+1)$, where $t \in [0, 1]$. First find the slope by differentiating the curve equation: \[ \frac{dy}{dx} = 2x \quad \Rightarrow \quad m = 2t \] Using the point-slope formula, write the equation of the tangent line: \[ y - (t^2 + 1) = 2t(x - t) \quad \Rightarrow \quad y = 2tx - 2t^2 + t^2 + 1 \quad \Rightarrow \quad y = 2tx - t^2 + 1 \]

Step 2: Find the vertical heights of the trapezium boundaries.
The trapezium is bounded horizontally by the vertical lines $x = 0$ and $x = 1$. Let us find the heights at these endpoints:
• At the left boundary $x = 0$: $y_1 = 2t(0) - t^2 + 1 = 1 - t^2$
• At the right boundary $x = 1$: $y_2 = 2t(1) - t^2 + 1 = 1 + 2t - t^2$

Step 3: Construct the area function of the trapezium.
The area of a trapezium with parallel vertical heights $y_1, y_2$ and a shared base width of $\Delta x = 1$ is: \[ A(t) = \frac{y_1 + y_2}{2} \cdot 1 = \frac{(1 - t^2) + (1 + 2t - t^2)}{2} = \frac{2 + 2t - 2t^2}{2} = 1 + t - t^2 \]

Step 4: Maximize the area function using optimization.
To find where the area reaches its maximum value, differentiate $A(t)$ with respect to $t$ and set it equal to zero: \[ A'(t) = 1 - 2t = 0 \quad \Rightarrow \quad t = \frac{1}{2} \] Since the second derivative is strictly negative ($A''(t) = -2 < 0$), this critical point represents a true local maximum. Now calculate the corresponding $y$-coordinate on the curve: \[ y = \left(\frac{1}{2}\right)^2 + 1 = \frac{1}{4} + 1 = \frac{5}{4} \] Thus, the tangent must be drawn at the coordinate point $\left(\frac{1}{2},\frac{5}{4}\right)$, corresponding to option (C).