# Math Help - Maximum values

1. ## Maximum values

in triangle PQR, PQ=(x+2)cm, PR=(5-x)cm and <QPR=30. the area(A) of the trangle is 1/4(10+3x-x^2)

the question i am struggling with is this; "use the method of completing the square to find the maximum value of A and the corresponding value of x."

i'd be grateful for someone to explain how its done and to also say what is done if it is a minimum value in case that another question asks me to find the minimum. thanks MHF.

2. Originally Posted by Tom G
in triangle PQR, PQ=(x+2)cm, PR=(5-x)cm and <QPR=30. the area(A) of the trangle is 1/4(10+3x-x^2)

the question i am struggling with is this; "use the method of completing the square to find the maximum value of A and the corresponding value of x."

i'd be grateful for someone to explain how its done and to also say what is done if it is a minimum value in case that another question asks me to find the minimum. thanks MHF.
$\frac14(10+3x-x^2)$

$\Rightarrow\frac14(\frac94 + 10 - \frac94 + 2\frac32 x-x^2)$

$\Rightarrow\frac14(\frac94 + 10 - (x - \frac32)^2)$

So you are subtracting from some number, when will it be maximum??
Ans: If you subtract minimum value. minimum of $(x - \frac32)^2$ is 0 and that is when $x=\frac32$

So maximum value of Area is $\frac{49}{16}$

3. Originally Posted by Tom G
in triangle PQR, PQ=(x+2)cm, PR=(5-x)cm and <QPR=30. the area(A) of the trangle is 1/4(10+3x-x^2)

the question i am struggling with is this; "use the method of completing the square to find the maximum value of A and the corresponding value of x."

i'd be grateful for someone to explain how its done and to also say what is done if it is a minimum value in case that another question asks me to find the minimum. thanks MHF.
If you go back to C1 notes you will find that:

$a(x-b)^2+c$
$a>0$

has minimum value $c$ at $x=b$

and

$-a(x-b)^2+c$
$a>0$

has maximum value $c$ at $x=b$