Prove that ax+bx+ay+by ≤ 300. Let $a,b,x,y$ be positive numbers satisfying:
$ax ≤ 100,  bx ≤ 100$,
$ay ≤ 100,  by ≤ 50$.
Prove that $ax+bx+ay+by ≤ 300$.
Can someone help me ?? 

There are three cases, according to the position of $a$ viz $b$
and $2b$.
Case 1. $a \leq b$.
Then 
$$
(a+b)(x+y) \leq (a+b)(\frac{100}{b}+\frac{50}{b})=150\frac{a+b}{b} \leq
150\frac{b+b}{b}=300.
$$
Case 2. $b \leq a \leq 2b$.
Then 
$$
(a+b)(x+y) \leq (a+b)(\frac{100}{a}+\frac{50}{b})=
150+\frac{100b}{a}+\frac{50a}{b}=
300-\frac{50(a-b)(2b-a)}{ab} \leq 300.
$$
Case 3. $2b \leq a$.
Then 
$$
(a+b)(x+y) \leq (a+b)(\frac{100}{a}+\frac{100}{a})=200\frac{a+b}{a} \leq
200\frac{a+\frac{a}{2}}{a}=300.
$$
