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Answer 1 · 2017-08-11T13:23:42

$13 +13 = 26$

$13xx13 = 169$

Lets arrange pairs of numbers which add to $26$ in order and find their product.

$0+26:" "rarr 0xx26 =0$
$1+25:" "rarr 1xx25 =25$
$2+24:" "rarr 2xx24=48$
$3+23:" "rarr 3xx23 =69$

$7+19:" "rarr 7xx19 =133$
$10+16" "rarr10xx16 =160$

We notice that as the numbers get closer, the difference between them gets less, but their product increases.

The smallest difference will be when the two numbers are equal and their product will therefore be the biggest.

$12+14:" "rarr 12xx14 = 168$
$13+13:" "rarr 13xx13 = 169$

Answer 2 · 2017-08-11T14:29:38

The numbers are $13$ and $13$

Let one number be $=x$

Then, the other number is $=26-x$

Let the product of the numbers be $=y$

So,

$y=x(26-x)=26x-x^2$

Differentiating with respect to $x$

$y'=dy/dx=26-2x$

The critical value is when $dy/dx=0$

$=>$ , $26-2x=0$

$=>$ , $x=13$

Let's make a variation chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaaaaa)$ $13$ $color(white)(aaaaaaa)$ $+oo$

$color(white)(aaaa)$ $y'$ $color(white)(aaaaaaaa)$ $+$ $color(white)(aaaa)$ $0$ $color(white)(aaaa)$ $-$

$color(white)(aaaa)$ $y$ $color(white)(aaaaaaaaa)$ $↗$ $color(white)(aaaa)$ $0$ $color(white)(aaaa)$ $↘$

There is a maximum at $x=13$

We confirm this by calculating the second derivative

$y''=(d^2y)/dx^2=-2$

As $y''<0$ , we conclude that when $x=13$ , there is a maximum value