Which of these graphs shows the area $A(c)$ as $c$ varies?

Which of these graphs shows the area $A(c)$ as $c$ varies?

Question

Shown below is a diagram of the square with vertices $(0, 0), (0, 1), (1, 1), (1, 0)$ and the line $y = x+c$. The shaded region is the region of the square which lies below the line; this shaded region has area $A(c)$.

Which of the following graphs shows $A(c)$ as $c$ varies?

Suggestion

What values of $c$ give a positive area?

When will the area be growing quickly and when slowly?

Which of the graphs given are in line with this?

Alternatively; can we come up with an explicit formula for $A(c)$, one that tells us what $A(c)$ is terms of $c$?

Could we consider the two cases $-1 \leq c \leq 0$ and $0 \leq c \leq 1$ separately?

What shape would the graph of such a formula be?

Solution

The area $A(c)$ is first positive once $c$ becomes greater than $−1$, which is when the line $y = x + c$ first touches the square.

As $c$ increases, the area increases more and more quickly (as more of the line $y = x + c$ lies within the square) until $c = 0$ (when $y = x$ is a diagonal of the square).

The area then increases more and more slowly until $c = 1,$ at which point $A(c)$ reaches its maximum value.

The graph with this shape is (a).

Algebraically, looking at the areas of triangles, when $-1 \leq c \leq 0$, we have $A(c) = \dfrac{1}{2}(1+c)^2,$ and when $0 \leq c \leq 1$, we have $A(c) = 1-\dfrac{1}{2}(1-c)^2 = c+\dfrac{1}{2} -\dfrac{c^2}{2}.$

Thus for $-1 \leq c \leq 0$, we have part of a vertex-down parabola, while for $0 \leq c \leq 1$, we have part of a vertex-up parabola.

The only possible answer is (a).