IIT-JEE 2002Question 1
A straight line $L$ with negative slope passes through the point $(8,2)$ and cuts the positive coordinate axes at points $P$ and $Q$. Find the absolute minimum value of $OP+OQ$, as $L$ varies, where $O$ is the origin.
Differentiation and its Application
Answer$18$
Solution
Concept. Write the line in intercept form; the point $(8,2)$ gives one condition on $a,b$. Bound $a+b$ by the Cauchy-Schwarz inequality. Formulas used. (1) Intercept form: $\frac xa+\frac yb=1$(2) Cauchy-Schwarz: $(a+b)\left(\frac{p^2}{a}+\frac{q^2}{b}\right)\ge(p+q)^2$, equality iff $\frac ap=\frac bq$Step 1 - Condition. By (1), with $a,b>0$: $\frac8a+\frac2b=1$.Step 2 - Bound. By (2) with $p=\sqrt8$, $q=\sqrt2$: $$a+b=(a+b)\left(\frac8a+\frac2b\right)\ge\left(\sqrt8+\sqrt2\right)^2=\left(3\sqrt2\right)^2=18.$$ Step 3 - Equality. $\frac{a}{\sqrt8}=\frac{b}{\sqrt2}$ gives $a=2b$; then $\frac{8}{2b}+\frac2b=\frac6b=1$, so $b=6$, $a=12$. The minimum of $OP+OQ$ is $18$. Answer: $18$