Question

Three numbers are given whose sum is 180 and the ratio of the first two is 1:2.If the product of the numbers is greatest,find the numbers.

Answer 1

Let the numbers be $x$, $y$ and $z$ then its given that $x+y+z=180$

$\frac{x}{y}=\frac{1}{z}$

$\Rightarrow y=2xx+y+z=180$

$\Rightarrow y=2x$ and $z=180-y-x$

$=180-2x-x$

$=180-3x$

Let $P$ denote product of the numbers

Then $P=xyz$

$P=x\times 2x\times (180-3x)$

$={2x}^2(180-3x)$

$P={360x}^2-{6x}^3$

$\frac{dP}{dx}=720x-18x^2$

$\frac{d^{2}P}{{dx}^2}=720-36x$

For maximum or minimum value of $P$

we must have

$\frac{dP}{dx}=0$

$\Rightarrow 720x-18x^2=0$

$x=40$

${\left(\frac{d^{2}P}{{dx}^2}\right)}_{x=40}=720-360\times 40<0$

Hence $P$ is maximum when $x=40$

Now, $x=40$ $y=2x=2\times =80$ $z=180-3x=180-3\times 40=180-120=60$.

Numbers are $40,80$ and $60$.