Basic inequality if the perimeter of a right triangle is t, then the maximum area of the right triangle is t

Basic inequality if the perimeter of a right triangle is t, then the maximum area of the right triangle is t


Let the area of an isosceles right triangle be the largest! Let the length of the right side be x, then the length of the hypotenuse is t-2x. From the Pythagorean theorem, we can get: 2x & sup2; = (t-2x) & sup2;, X1 = [2T + √ (2t & sup2;)] / 2 [rounding off], X2 = [2T - √ (2t & sup2;)] / 2. The largest area is: X & sup2; = {[2T - √ (2t & sup2;)] /



Why can mean inequality be used to find the maximum area of a right triangle when its perimeter is 10
If one side of the right angle is x and the other side is y, x + y is not a fixed value, x + y + √ (x ^ 2 + y ^ 2) = 10 ≥ 2 √ (XY) + √ (2XY)
Why can mean inequality be used


If x > 0, Y > 0, then x + Y > = 2 √ (XY) can be used
And (x ^ 2 + y ^ 2) > = 2XY can be used at any time, no matter x > 0 or Y > 0



If the radius of a is 10m, when the radius decreases x (m), the area of the circle decreases y (M & sup2;). Y is a function of X. write out the analytic formula of the function to fit it


y=π(10-x)^2 (x



If the radius of a circle is 4cm, and the radius increases by xcm, the area of the circle increases by y2cm. Find the functional relationship between Y and X


y = π(x+4)^2-π*4^2 = πx^2+8πx+16π-16π = πx^2+8πx