Question

If a,b,c,d are positive real numbers (a > c), then value of d that minimize the expression $\sqrt{a^2+d^2}+\sqrt{(b-d{)}^2+c^2}$ is / are

• A. $\frac{a+c}{ab}$
• B. $\frac{ac}{a+b}$
• C. $\frac{a+b}{ac}$
• D. $\frac{ab}{a-c}$

Answer 1

The correct option is A $\frac{a+c}{ab}$

$f\left(d\right)=\sqrt{a^2+d^2}+\sqrt{{(b-d)}^2+c^2}$

$f\left(d\right)isminimumwhere{f}^{\prime }\left(d\right)=0$

$f^{\prime }\left(d\right)=\frac{d}{\sqrt{a^2+d^2}}+\frac{2d-2b}{2\sqrt{b^2+d^2+c^2-2bd}}=0$

$\Rightarrow d.\sqrt{b^2+d^2+c^2-2bd}+(b-d)\sqrt{a^2+d^2}=0$

$d^2(b^2+d^2+c^2-2bd)=(b^2+d^2-2bd)(a^2+d^2)$

$b^2{d}^2{+d}^4+d^2{c}^2-2b{d}^3=a^2{d}^2+a^2{b}^2-2a^{2}bd+d^4+b^2{d}^2-2b{d}^3$

$d^2(c^2-a^2)+\left(2a^{2}b\right)d-a^2{b}^2=0$

$d=\frac{-2a^{2}b\pm \sqrt{4a^4{b}^2+4\left(a^2{b}^2\right)(c^2-a^2)}}{2(c^2-a^2)}$

$=\frac{-2a^{2}b\pm 2abc}{2(c^2-a^2)}$

$d=\frac{-ab(a+c)}{c^2-a^2},\frac{-ab(a-c)}{c^2-a^2}$

$d=\frac{ab}{a-c},\frac{ab}{a+c}$

Hence, the answer is $\frac{ab}{a-c}$