In ΔABC: D is 'n punt op AB, E is 'n punt op AC en F is 'n punt op BC zodat dat DECF 'n parallelogram is - NSC Mathematics - Question 10 - 2018 - Paper 1

To calculate the area of the parallelogram DECF given that the height AG = 1 unit and the base BC = (5 - t) units:

The area $A$ of a parallelogram is given by the formula:

$A = ext{base} imes ext{height}$.

Substituting the values:

$A(t) = (5 - t) imes 1 = 5 - t$.

To find the maximum area, we differentiate $A(t)$ with respect to $t$ and set it to zero:

$A'(t) = -1$.

Thus, $A$ is maximized when $t$ is minimized. Since $t$ cannot exceed 5, we evaluate:

$A_{ ext{max}} = 5$.

Putting this into context from the marking scheme, we also observe the specific values related to areas:

A(t) = rac{6}{25}(5-t) + rac{6}{5}.

From the calculations, we can derive:

For $t = 0$:

$A(0) = (5) = 5$

For $t = 5$:

A(5) = rac{30}{12} = 2.5.

Hence, the area can vary based on the values of t, and thus indicate potential maxima or minima depending on constraints provided.