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.
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.
Hello, geton!
$\displaystyle \text{In }\Delta ABC,\;AB \,= \,2 - x,\;BC \,= \,x + 1,\;\angle ABC = 120^o$
a) Show that: .$\displaystyle AC^2 \:= \:x^2 - x +7$
b) Find the value of $\displaystyle x$ for which $\displaystyle AC$ has a minimum value.
You already have: .$\displaystyle AC \:=\:(x^2-x+7)^{\frac{1}{2}}$
Minimize the function: .$\displaystyle AC' \;=\;\frac{1}{2}(x^2-x+7)^{-\frac{1}{2}}(2x-1)$
We have: .$\displaystyle \frac{2x-1}{2\sqrt{x^2-x+7}} \:=\:0\quad\Rightarrow\quad 2x-1 \:=\:0$
Therefore: .$\displaystyle x \:=\:\frac{1}{2}$