The given figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 30 units long. So the first thing we're going to do is we want to come up with an equation for line AB. And if we know that this is an isosceles right triangle, then if we dropped an altitude to the base, we know that that would split these two angles equally and a 45 and 45 forming another right triangle, with this already being 45 from the original isosceles. So if the distance from the origin to A is 15, then the distance from the origin to B is also 15. So two points on this line would be zero, 15 and 15 zero. If we know two points on a line, we can always find the slope. So the change of YS over the change of X we get a slope and -1 using our point slope formula we get y -, 15 equaling -1 times the quantity X -, 0. Remember, it doesn't matter which of the two points we used. I just happened to use the zero 15 or the Y intercept point. So Y equal negative X + 15. So now we know that this point over here is going to be X amount over and negative X + 15 up. Now that only gives us half. That gives us this portion here. So if we look at the whole base, we know that we need another X distance or X + X times the height of negative X + 15. X plus X gives us 2X. If we distribute the 2X, we get negative two X ^2 + 30 X. Finding the derivative, we get negative four X + 30. Setting the derivative equal to 0 because we want to find the Max or the min, we get X equaling 15 halves. But remember X was only half of the whole base. So 2X or 2 * 15 halves would give us a base of 15 units. Our height is 15 minus X 15 -, 15 halves, 30 halves -15 halves, 15 halves units for the height. So our base is 15 and our height is 15 halves. Finding the area, we're just going to multiply 15 by 15 halves and get 225 halves square units for the actual area. So the dimensions of the rectangle having the largest area are going to be 15 units by 15 halves units.