Law of Cosines and to Find Extremum by Completing Square

In $\displaystyle \Delta ABC$, $\displaystyle AB=(5-x)$ cm, $\displaystyle BC=(4+x)$ cm, $\displaystyle \angle AsBC=120{}^\circ$ and $\displaystyle AC=y$ cm.

(a)     Show that $\displaystyle {{y}^{2}}={{x}^{2}}-x+61$.

(b)     Find the minimum value of $\displaystyle {{y}^{2}}$, and give the value of $\displaystyle x$ for which this occurs.

Solution
$\displaystyle AB=(5-x)$ cm,
$\displaystyle BC=(4+x)$ cm,
$\displaystyle \angle ABC=120{}^\circ$
$\displaystyle AC=y$ cm.

(a)      By the law of cosines,

$\displaystyle A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2\cdot AB\cdot AC\cos (\angle ABC)$

$\displaystyle {{y}^{2}}={{(5-x)}^{2}}+{{(4+x)}^{2}}-2(5-x)(4+x)\cos 120{}^\circ$

$\displaystyle {{y}^{2}}=25-10x+{{x}^{2}}+16+8x+{{x}^{2}}+2(5-x)(4+x)\left( {\frac{1}{2}} \right)$

$\displaystyle {{y}^{2}}=41-2x+2{{x}^{2}}-{{x}^{2}}+x+20$

$\displaystyle {{y}^{2}}={{x}^{2}}-x+61$

$\displaystyle \therefore {{y}^{2}}={{x}^{2}}-x+\frac{1}{4}+61-\frac{1}{4}$

$\displaystyle \therefore {{y}^{2}}={{\left( {x-\frac{1}{2}} \right)}^{2}}+60.75$

Since $\displaystyle {{\left( {x-\frac{1}{2}} \right)}^{2}}\ge 0\$ for all $\displaystyle x\in R$,

$\displaystyle {{\left( {x-\frac{1}{2}} \right)}^{2}}+60.75\ge 60.75$

$\displaystyle \therefore {{y}^{2}}\ge 60.75$.

Therefore the minimum value of $\displaystyle {{y}^{2}}$ is $\displaystyle 60.75$ and this value occurs when $\displaystyle x=\frac{1}{2}$.