Two positive numbers x and y whose sum is 25 and the product $x^3y^2 $ is maximum are : |
$x=10, y = 15$ $x=15, y = 10$ $x=12, y = 13 $ $x=16, y = 9 $ |
$x=15, y = 10$ |
The correct answer is Option (2) → $x=15, y = 10$ $x+y=25⇒y=25-x$ and, $x^3y^2=x^3(25-x)^2$ $=x^3(625+x^2-50x)$ $=625x^3+x^5-50x^4$ $\frac{d(625x^3+x^5-50x^4)}{dx}=1875x^2+5x^4-200x^3$ for critical point, $⇒1875x^2+5x^4-200x^3=0$ $⇒x^4-40x^3+375x^2=0$ $⇒x^2(25-x)(75-5x)=0$ Since $x>0$ and $y>0$, $75-5x=0$ $5x=75⇒x=15$ and $y=10$ |