The diagonals of a convex quadrilateral are mutually perpendicular. The sum of the lengths of the diagonals is 12. We want to find the maximum possible area of such a quadrilateral. Let us denote by x and y the lengths of the two diagonals. Then the area of the quadrilateral is the following function of x and y:

Maximum area =18

The area of the quadrilateral as a function of x and y = xy/2

Step-by-step explanation:

As the quadrilateral has mutually perpendicular diagonals, it is a rhombus. The area of a rhombus is denoted by the formula xy/2.

So, the area of the quadrilateral is xy/2.

As the sum of the sides is 12

x + y = 12

y = 12- x

So, the area, as a function of x alone, becomes x(12- x)/2

To find the maximum area, we find the derivative of the area function with respect to x and equate it to 0.

d/dx(x(12 - x)/2) = 0

d/dx(12x - x^2) = 0

12- 2x = 0

x = 6

The maximum area will then be 6(12- 6)/2 = 36/2 = 18

So, the maximum area is 18.

ur mom gae no homo

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

we now create space between the x and y terms with parenthesis.

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

we add the squares to both sides.

simplify.

and write the quadratics in factored form.

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.