00:04
Now in this question, we look at a rectangular page, right, which let's say something like this, right? i suppose this is a page we look at, right? and i suppose this is x and this is y, okay? and something we need to find out.
00:19
Now, the margins, let me put, let's look at the margins.
00:24
So suppose some of the margins, right? so the principal area, of course, will be actually, we're being by this, right? now the margins at the top and the bottom would be one inch, right? so this part is a one, and of course, this part is also one way.
00:38
And the left side, on one side, it's one inch.
00:44
So this is all one inch, but this side actually is two inches, right? so the printable areas here.
00:50
So what is the printable area? we can work it out, right? so the print of a printable area, of course, is giving by the lens of this, multiple by the lens of this, right? now you can find out the lens of this, right? that's of course given by x minus one and two, right? so that's x minus 3 and multiply by the length of this side, right? that's going by y minus 1 and the 1, 1, 1, 1, 1 minus 2, right? so that would be the area.
01:15
Now, the total area of this has to, according to the question has to be 84, right? and the question is, you're asked to find the minimal paper you need to use, right? that's x, y, right? that's minimum paper we need to find, right? so that's xy.
01:35
And of course, from this, what we want to do is to, you know, basically we just need different shades, right? we just need different shades as a function of x, right? and then we can find, because we came from this, we can actually express y in terms of x.
01:51
So let's do that.
01:52
X times now y from this equation, we can find a y, right? y is given by 84 over x minus 3, right, and plus 2, right? so this is the function of x.
02:02
And now to find the, to minimize the s, we just need to take the derivative function of this s, right? i call that derivative function s prime x, right? so you do the definition, you find two terms.
02:17
One is, of course, 84 over x minus 3 plus 12, plus 2, right? and we have another term, which of course, it comes for a different of this part.
02:26
That's plus x times.
02:28
Because this x minus 3 is a bottom, you get a minus sign here.
02:32
And you get x minus 3 squared and i get 84 here right so that will be it and of course in the end you let's we have to set this to the zero okay and then you get the x and of course to do this we can we can make it a bit better 84 minus over x3 right and we can take out these two terms and that will become one minus 84 over x minus 3 right so that would be it and plus 2 and this to bit 0 right and so that's basically the equation you have to solve, right? and of course, we can solve this equation.
03:09
What we want to do is just to multiply both sides by x minus 3 squared, right? and then you'll find this to be actually given by 84 times x minus 84x, right? and then of course, plus 2 times x minus 3 squared, right? i can become zero.
03:35
So this equation, you can write a fast.
03:37
Now, this is a standard equation, which you can easily solve...