If y is a real number,what is the difference in the maximum and minimum values obtained by y+5/y^2+5y+25.Please explain in detail how we go about solving this problem.

The given expression is $$\frac{y+5}{y^2+5y+25}$$. Let's call it E.
This can be rewritten as,
E = $$\frac{y+5}{y^2+10y+25-5y}$$
E = $$\frac{y+5}{(y+5)^2-5y}$$
E = $$\frac{1}{(y+5)-\frac{5y}{y+5}}$$
Let's assume x = y+5
E = $$\frac{1}{x-\frac{5(x-5)}{x}}$$
E = $$\frac{1}{x-5+\frac{25}{x}}$$
E = $$\frac{1}{(x+\frac{25}{x})-5}$$
The expression $$x+\frac{25}{x}$$ can be differentiated to find its local minimum and maximum.
Differentiating the expression we get, $$1-\frac{25}{x^2}=0$$ => x = 5 or -5.
The expression E is minimum when x = -5 and maximum when x = 5.
Min(E) = $$\frac{1}{(x+\frac{25}{x})-5}$$ = $$-\frac{1}{15}$$
Max(E) = $$\frac{1}{(x+\frac{25}{x})-5}$$ = $$\frac{1}{5}$$
The difference between the max and min values = $$\frac{1}{5}-(-\frac{1}{15}) = \frac{4}{15}$$