Question

Find two positive numbers x and y such that their sum is 35 and the product $x^2{y}^5$ is maximum.

Answer 1

Let the numbers be x and y and $p=x^2{y}^5,thenx+y=35$
$\Rightarrow x=35-y,\therefore p=(35-y{)}^2{y}^5$
On differentiating twice w.r.t. y, we get
$\frac{dP}{dy}=(35-y{)}^{2}5y^4+y^{5}2(35-y\left)\right(-1)=y^4(35-y\left)\right[5(35-y)-2y]=y^4(35-y\left)\right(175-5y-2y\left)y^4\right(35-y\left)\right(175-7y)=(35y^4-y^5\left)\right(175-7y)and\frac{d^P}{d{y}^2}=(35y^4-y^5\left)\right(-7)+(175-7y\left)\right(4\times 35\times y^3-5y^4)=-7y^4(35-y)+7(25-y)\times 5y^3(28-y)=-7y^4(35-y)+35y^3(25-y\left)\right(28-y)$
For maxima put $\frac{dP}{dy}=0$
$\Rightarrow y^4(35-y)(175-7y)=0\Rightarrow y=0,35-y=0,175-7y=0\Rightarrow y=0,y=25,y=35$
When y=0, x=35-0=35 and the product $x^2{y}^5$ will be 0.
When y=35 and x=35-35=0. This will make the product $x^2{y}^5$ equal to 0.
$\therefore$ y=0 and y=35 cannot be the possible value of y.
When y=25,
${\left(\frac{d^{2}P}{d{y}^2}\right)}_{y=25}=-7\times (25{)}^4\times (35-25)+35\times (25{)}^3\times (25-25)(28-25)=-7\times 390625\times 10+35\times 15625\times 0\times 3=-27343750+0=-27343750<0$
$\therefore$ By second derivative test, P will be he maximum when y=25 and x=35-25=10.
Hence, the required numbers are 10 and 25.