Answer
$600 \sqrt {7}$ mi/h
Work Step by Step
Reference to the fig in triangle APB
$z^2=x^2+y^2-2xy\cos120^\circ$
$z^2=x^2+y^2-2xy(-\frac{1}{2})$
$z^2=x^2+y^2+xy$ ................................. eq (1)
Putting $x=2$mi, $y=4$mi in equation (1)
$z^2=2^2+4^2+8=28 $ Or
$z=2\sqrt {7}$ mi
Taking derivative with respect to t of equation (1)
$\frac{d(z^2)}{dt}=\frac{d(x^2)}{dt}+\frac{d(y^2)}{dt}+\frac{d(xy)}{dt}$
$\frac{d(z^2)}{dz} \frac{dz}{dt}=\frac{d(x^2)}{dx} \frac{dx}{dt}+\frac{d(y^2)}{dy} \frac{dy}{dt}+\frac{d(xy)}{dt}$
$2z\frac{dz}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}+x\frac{dy}{dt}+y\frac{dx}{dt}$
$2z\frac{dz}{dt}=(2x+y) \frac{dx}{dt}+(2y+x)\frac{dy}{dt}$ ...................... eq (2)
Putting $ x=2$ mi, $y=4$ mi, $z=2\sqrt {7}$ mi, $\frac{dx}{dt}=600$ mi/h, $\frac{dy}{dt}=1200$ mi/h in equation (2)
$2\times 2\sqrt {7}\frac{dz}{dt}= (2\times 2+4)\times 600+ (2 \times4+2)\times1200$
$2\times 2\sqrt {7}\frac{dz}{dt}= (2\times 2+4)\times 600+ (2 \times4+2)\times1200$
$\frac{dz}{dt}= \frac{16800}{4\sqrt {7}}= \frac{4200}{\sqrt {7}}$
$\frac{dz}{dt}= \frac{4200}{\sqrt {7}} \times \frac{\sqrt {7}}{\sqrt {7}}= \frac{42 00\sqrt {7}}{(\sqrt {7})^2}= \frac{4200 \sqrt {7}}{7}=600\sqrt {7}$ mi/h
