ADEC is a rectangle with |C| = 7 m and |D| = 2 m, as shown. B is a point on |AC| such that |AB| = 5 m. P is a point on |DE| such that |DP| = x m. (i) Let $f(x) = |PA|^2 + |PB|^2 + |PC|^2$. Show that $f(x) = 3x^2 - 24x + 86$, for $0 \leq x \leq 7, x \in \mathbb{R}$. (ii) The function $f(x)$ has a minimum value at $x = k$. Find the value of $k$ and the minimum value of $f(x)$.

Leaving Cert Mathematics Area & Volume: ADEC is a rectangle with |C| = 7

ADEC is a rectangle with |C| = 7 m and |D| = 2 m, as shown - Leaving Cert Mathematics - Question b - 2014

Question b

ADEC is a rectangle with |C| = 7 m and |D| = 2 m, as shown. B is a point on |AC| such that |AB| = 5 m. P is a point on |DE| such that |DP| = x m. (i) Let $f(x) = |P... show full transcript

Worked Solution & Example Answer:ADEC is a rectangle with |C| = 7 m and |D| = 2 m, as shown - Leaving Cert Mathematics - Question b - 2014

Step 1

Let $f(x) = |PA|^2 + |PB|^2 + |PC|^2$

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Answer

To find $f(x)$ , we first calculate the distances:

Calculate the length |PA|:
- |PA| = |PD| + |DA| = x + 2
Calculate the length |PB|:
- Since B is at (5, 0) and P is (x, 2), using the distance formula: $|PB| = \sqrt{(5 - x)^2 + (0 - 2)^2} = \sqrt{(5 - x)^2 + 4}$
Calculate length |PC|:
- |PC| = |PE| + |EC| = (7 - x) + 2 = 9 - x

Now substituting into the function:

$f(x) = (x + 2)^2 + \left( \sqrt{(5 - x)^2 + 4} \right)^2 + (9 - x)^2$

$f(x) = (x^2 + 4x + 4) + ((5 - x)^2 + 4) + (9 - 2x + x^2)$

Expanding and simplifying:

$= x^2 + 4x + 4 + (25 - 10x + x^2 + 4) + (81 - 18x + x^2)$ $= 3x^2 - 24x + 86$

Step 2

The function $f(x)$ has a minimum value at $x = k$

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Answer

To find the minimum value of $f(x)$ , we differentiate it:

Find the first derivative: $f'(x) = 6x - 24$
Set the first derivative to zero to find the critical points: $6x - 24 = 0 \implies x = 4$
Confirm that it is a minimum by checking the second derivative: $f''(x) = 6 > 0$ This confirms a local minimum at $x = 4$ .
Find the minimum value of $f(x)$ : $f(4) = 3(4)^2 - 24(4) + 86$ $= 48 - 96 + 86 = 38$

Thus, $k = 4$ and the minimum value of $f(x)$ is 38.

ADEC is a rectangle with |C| = 7 m and |D| = 2 m, as shown - Leaving Cert Mathematics - Question b - 2014

Worked Solution & Example Answer:ADEC is a rectangle with |C| = 7 m and |D| = 2 m, as shown - Leaving Cert Mathematics - Question b - 2014

Let $f(x) = |PA|^2 + |PB|^2 + |PC|^2$

The function $f(x)$ has a minimum value at $x = k$

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