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}$.
(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 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}$.
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