Is this a tricky question?

Let the two real numbers be x and y

Since, $x+y=100 \implies y = 100-x$

The product of the two numbers = $xy=x(100-x)=100x-x^2$

So, we need to maximise the function $100x-x^2$

Let $f(x)= 100x-x^2$

$f(x)$ is maximised when ${f}'(x)=0$

So, $100 - 2x = 0 \implies x = 50$

Thus, x = y = 50 maximises it.

Is this a troll question?

No, this is an introductory Calculus Question

The product of two natural positive numbers is the area of a rectangle with those two numbers as sides. Since max: value of a rectangle is possible when it is a sqare, the sides have to ve 50 each, and the area would be 2500

If you've ever done MATHCOUNTS, this is an essential trick. The product of two numbers with a given sum is maximized when both numbers are the same.

We proceed with the AM-GM inequality. If the numbers are $x$ and $y$ , we see that

$\dfrac{x+y}{2} \geq \sqrt{xy}$

The maximum value of the RHS will be achieved when equality is attained. Set both sides equal, and substitute $x+y=100$ . We get

$50=\sqrt{xy}$

Squaring,

$xy=\boxed{2500}$

Use AM greater than or equal to GM

(x+y)/2 is greater than or equal to (xy)^1/2

By substituting x+y as 100, We can get xy is less than or equal to 2500

(Sorry for using words..I dont know how to put greater than equal to sign)

Let the two positive real numbers be $x$ and $y$ . By $\text{AM-GM}$ , $\frac{x+y}{2} \geq \sqrt{xy}$ $\implies \sqrt{xy} \leq \frac{100}{2}=50$ $\implies xy \leq 50^2=\boxed{2500},$ with equality iff $x=y=50$ .

Well an easier approach would be to know that for a given sum formed by two real no., the maximum product is achieved, when the difference between the two no.s is the smallest.for example.lets take a no. say 16.clearly its an even no.,so it can be divided by 2, and so the two no.s(8,8) would have the least difference i.e, 0 when compared to any other pair(say 7,9) and so the maximum product would be 8*8=64.

I discovered it while learning the factorisation of a trinomial,.where splitting of the terms is required I think you should get it..though i have explained it too much briefly:)

Let the two positive numbers be c and d

Now, the arithmetic mean of c and d = $\frac{c+d}{2}$

And the geometric mean of c and d= $\sqrt{cd}$

AM >= GM $\frac{c+d}{2}$ >= $\sqrt{cd}$

$\frac{100}{2}$ >= $\sqrt{cd}$

$\sqrt{cd}$ <= 50

cd <= $50^{2}$

cd <= 2500

therefore the maximum possible product is 2500.

Instead of doing the calculus I just thought

$2\times98$ is definitely > $1\times 99$

That means $50\times 50$ will be the greatest but I thought it has to be 2 distinct numbers so I went with $49\times51$ first

Let the numbers be $a,b$ .By the AM-GM Inequality,we have: $\frac{a+b}{2}\geq \sqrt{ab}$ Putting in $a+b=100$ , we get: $\frac{100}{2}\geq \sqrt{ab}$ $50\geq \sqrt{ab}$ Squaring both sides,we get: $\boxed{2500\geq ab}$

• Let the two real numbers be a and b
• a+b=100 (given)..........(1)
• now as;
• AM >= GM
• (a+b)/2 >= (ab)^(1/2)
• 100/2>=(ab)^(1/2)
• 50>=(ab)^(1/2)
• now by squaring both side we get...
• 2500>=ab
• so the max value of ab is 2500.

Since both the numbers are positive...And Since for the positive numbers their arithmetic mean is always greater than or equal to their geometric mean...... Let a and b be the two numbers.. This implies (a+b)/2 >= (a*b)^0.5....... But since the product is to be maximum... This implies that equality holds.... Therefore both the numbers are equal.. But since their sum is 100 so both numbers are 50 each.... Hence their product is 2500.

It can be simply solved using AM-GM inequality
Let the two number number be $a,b$

$\frac {a+b}{2} \ge \sqrt{ab}$
$\rightarrow 50 \ge \sqrt{ab}$
$\rightarrow 2500 \ge ab$

$2500 = ab$ if and only if $a = b = 50$

So $ab = \boxed{2500}$

• a+b=100 so (a+b)/2 = 50 so (50)^2 = 2500

A very simple approach to this, not using Maxima and Minima or AM-GM inequality would be

Step 1:Take the numbers as 50+x and 50-x Step 2:Product of them is 50^2 - x^2

Since x=0 is the minimum value for x, we get the product as maximum(subtracting the minimum makes expression maximum)

Therefore product=50^2=2500

maximum product will only happen if a = b or b = a 100 = 2a a = 50 maximum P = 50 ( 50 ) = 2500

using AM-GM inequalities, (x+y)/2 >(and equal to) √xy,
substitute the given value, we get 50 >(and equal to) √xy
therefore, the maximum value of xy is 50^2 = 2500

The solution is yes, it is in fact a troll question.

it is given that a+b=100---- (1) . Let ab =y------ (2) (which is to be maximized). differentiate (1) and (2) wrt b. we get... da/db = -1---- (3) and da/db (b) +a =dy/db-----(4) sub,,. 3 in 4 and we get , a-b = dy/db .. since it has to be maximized, a-b=0----- (5). By solving eqn 4 and 5 we get a=50 and b=50 whose multiplication product is 2500.

This a simple question from maxima and minima. let, x+y=100 so that y=100-x xy=x(100-x) f(x)=100x-x^2. now find f'(x). get the value of x. substitue it in f(x) and get the max possible product!

50+50=100 50 50=2500 49+51=100 49 51=2499 That means when you make any addends less than 50, the product will become lesser.

Let x equal the first number, then 100-x is equal to the second number. The product is then (100-x)x = -x^2+100x which for convienience I will call f(x). Because this polynomial is of degree two it will have only one vertex. Then there is only one point on f(x) where the slope of the tangent is zero, namely the vertex. This sets up a nice equation 0=d/dx[f(x)]. This is just d/dx[-x^2+100x]=0 -> -2x+100=0 -> x=50. Knowing the x coordinate of the vertex is 50, we can find the y coordinate by finding f(50) = 5000-50^2 =5000-2500 = 2500. This is therefore the max of this equation, so our answer is 2500.//

Let x and y be the two numbers Now since x+y=100 Therefore x=100-y Now Let f(x)= xy therefore f(x)= (100-x)x Thus f(x) = 100x- $x^{2}$ Now f'(x) = 100- 2x ..(1) now For extreme values f '(x) = 0 thus From (1) X=50 so y =100-X=50 Thus x=50 and y=50 Maximises it Thus Product of two numbers =50×50= 2500

the simple way is we know the sum of the two numbers are 100 right? so we have to find the maximum possible product of the two numbers by multiplication .So try multiplying any number with another but it wont go over 50x50 that comes to 2500 so the answer is 2500..