Question

y=x(c-x) where c is constant.find maximum value of y

Answer 1

Answer:

Maximum value of y is c^2/4

Step-by-step explanation:

We have been given that

$y=x(c-x)$

Let us simplify this equation using distributive property a(b+c)=ab+ac

$y=cx-x^2\\y=-x^2+cx$

It represents a downward parabola. And we know that for a downward parabola, the vertex is the maximum point.

x-coordinate of vertex is $x=-\frac{b}{2a}\\\\x=-\frac{c}{-2}\\\\x=\frac{c}{2}$

y-coordinate of vertex is

$y=c\times\frac{c}{2}-(\frac{c}{2})^2\\\\y=\frac{c^2}{4}$

Hence, maximum value of y is c^2/4