There is a problem from sine & cosine chapter

**geton** · Jan 4th 2008, 07:44 PM

In ∆ABC, AB = (2 – x) cm, BC = (x + 1) cm and angle ABC = 120˚:

a) Show that AC^2 = x^2 – x +7.
b) Find the value of x for which AC has a minimum value.

I’ve done 1st part. But how can I solve 2nd part. Please help me.

**Soroban** · Jan 4th 2008, 07:57 PM

Hello, geton!

$\text{In }\Delta ABC,\;AB \,= \,2 - x,\;BC \,= \,x + 1,\;\angle ABC = 120^o$

a) Show that: . $AC^2 \:= \:x^2 - x +7$

b) Find the value of $x$ for which $AC$ has a minimum value.

You already have: . $AC \:=\:(x^2-x+7)^{\frac{1}{2}}$

Minimize the function: . $AC' \;=\;\frac{1}{2}(x^2-x+7)^{-\frac{1}{2}}(2x-1)$

We have: . $\frac{2x-1}{2\sqrt{x^2-x+7}} \:=\:0\quad\Rightarrow\quad 2x-1 \:=\:0$

Therefore: . $x \:=\:\frac{1}{2}$

**Krizalid** · Jan 4th 2008, 08:14 PM

Without differentiation, of course

$2 \cdot \overline {AC} = \sqrt {4x^2 - 4x + 28} = \sqrt {\left( {2x - 1} \right)^2 + 27} .$

The rest follows.

**geton** · Jan 4th 2008, 08:33 PM

Originally Posted by Krizalid

Without differentiation, of course

$2 \cdot \overline {AC} = \sqrt {4x^2 - 4x + 28} = \sqrt {\left( {2x - 1} \right)^2 + 27} .$

The rest follows.

You mean $\sqrt {\left( {2x - 1} \right)^2 + 27} = 0$

But in this way how can I get x = 1/2?

**Isomorphism** · Jan 4th 2008, 09:41 PM

Originally Posted by geton

You mean $\sqrt {\left( {2x - 1} \right)^2 + 27} = 0$

But in this way how can I get x = 1/2?

Since $\forall x \in \mathbb{R}, (2x-1)^2 \geq 0$ , we have $\sqrt{\left( {2x - 1} \right)^2 + 27} \geq \sqrt{27}$
with equality at 2x-1 = 0.
Thus min is achieved at $x = \frac12$

**geton** · Jan 4th 2008, 11:11 PM

Thanks.

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