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Question: If x and y are positive integers and \({{x}^{2}}+{{y}^{2}}=1800\), then maximum value of x+y is? A...

If x and y are positive integers and x2+y2=1800{{x}^{2}}+{{y}^{2}}=1800, then maximum value of x+y is?
A. 60
B. 52
C. 64
D. 48

Explanation

Solution

Hint: Before solving this problem, You need to know the relation A.M.≥G.M.\text{A}\text{.M}\text{.}\ge \text{G}\text{.M}\text{.} to get to the answer in the shortest possible manner. But before applying the relation convert the equation x2+y2=1800{{x}^{2}}+{{y}^{2}}=1800 to suitable form by using some basic algebraic formula and find the maximum value of x+y.

Complete step-by-step answer:
Given,
x2+y2=1800...........(i){{x}^{2}}+{{y}^{2}}=1800...........(i)
To maximise: x + y
Now, we know:
(a+b)2=b2+a2+2ab{{\left( a+b \right)}^{2}}={{b}^{2}}+{{a}^{2}}+2ab
Rearranging to suitable form;
(a+b)2−2ab=b2+a2{{\left( a+b \right)}^{2}}-2ab={{b}^{2}}+{{a}^{2}}
Using the mentioned formula in equation (i) we get;
x2+y2=1800{{x}^{2}}+{{y}^{2}}=1800
⇒(x+y)2−2xy=1800\Rightarrow {{\left( x+y \right)}^{2}}-2xy=1800
⇒(x+y)2=1800+2xy\Rightarrow {{\left( x+y \right)}^{2}}=1800+2xy
We know;
k2=±k\sqrt{{{k}^{2}}}=\pm k
Applying;
x+y=±1800+2xyx+y=\pm \sqrt{1800+2xy}
Now, as x and y are positive there sum must also be positive:
⇒x+y=1800+2xy.....................(ii)\Rightarrow x+y=\sqrt{1800+2xy}.....................(ii)
Now, we have to maximise x+y.
And from equation (ii) it is clear that for x+y to be maximum, xy should be maximum.
So, using the relation A.M.≥G.M.\text{A}\text{.M}\text{.}\ge \text{G}\text{.M}\text{.} we get;
A.M. of x and y = x+y2\dfrac{x+y}{2}
G.M. of x and y= xy\sqrt{xy}
∴x+y2≥xy\therefore \dfrac{x+y}{2}\ge \sqrt{xy}
And for xy to be maximum, xy\sqrt{xy} must be maximum.
So, G.M. is maximum when it is equal to A.M. and it is possible only if x and y are equal.
Therefore, we can say that x+y is maximum when x and y are equal.
Using this result in equation (i);
x2+y2=1800{{x}^{2}}+{{y}^{2}}=1800
⇒x2+x2=1800\Rightarrow {{x}^{2}}+{{x}^{2}}=1800
⇒2x2=1800\Rightarrow 2{{x}^{2}}=1800
⇒x2=900\Rightarrow {{x}^{2}}=900
⇒x=±900\Rightarrow x=\pm \sqrt{900}
⇒x=± 30\Rightarrow x=\pm \text{ }30
And it is mentioned that x is positive so, x = 30.
x=y=30x=y=30 for maximum value of x+y.
Therefore, maximum value of x+y is 30+30=6030+30=60
Hence, the answer is option A) 60.

Note: This method is only possible to use when all the variables involved in the relation of A.M.≥G.M.\text{A}\text{.M}\text{.}\ge \text{G}\text{.M}\text{.}are positive and the relation can be further extended as A.M.≥G.M.≥H.M.\text{A}\text{.M}\text{.}\ge \text{G}\text{.M}\text{.}\ge \text{H}\text{.M}\text{.} . In such questions you can also give a try to the method of Derivatives for maximising an expression as well. However, that might get a bit complicated but if you can solve that you will surely get the answer.