Rectangle in Semicircle

Let the centre of the circle be $(0,0)$ . The rectangle touches the circumference of the semicircle at $(x,y)$ .

Given that

$x^2+y^2=6^2$

We need to find the maximum value of its perimeter, $4x+2y$ .

By Cauchy-Schwarz Inequality,

$(x^2+y^2)(4^2+2^2)\geq(4x+2y)^2$

$36\times20\geq(4x+2y)^2$

$4x+3y\leq12\sqrt5$ .

Hence the maximum value of its perimeter is $\boxed{12\sqrt5}$ .

Draw a semi circle. Draw x and y axes with origin at the center and place the rectangle symmetrically with the height as y and the length will be 2x. The perimeter will be P=4x + 2y. Draw a line from O to the point P on any of the two points on the circle. This will have a radius of 6. By Pythagorean theorem, x^2 + y^2 = 36. Solving for y, y = sqrt of 36-x^2. Substitute in P. P=4x + 2(sqrt of 36-x^2). Differentiate P with respect to x and equate to zero solving x=12/sqrt of 5. Substitute in the Pythagorean formula to get y=6/sqrt of 5. Substitute x and y in P to get 12*sqrt of 5.

let O be the centre of the circle. if rectangle's AB is on the diameter then OD=OC=6 (as 6 be the radius of the semicircle). let $\alpha$ be the angle between OC and BC . then BC=6 $\cos \alpha$ and AB=2 $\times 6$ $\sin \alpha$ =12 $\sin \alpha$ . so the perimeter of the rectangle will be 2 $\times 6$ $\cos \alpha$ +2 $\times 12$ $\sin \alpha$ . then using the derivative process for maxima of a function we will get the perimeter $\boxed{12\sqrt{5}}$ .

honestly i just took a guess...i have not been good at math and admittedly, i have probably never have or found very few situations where i have had to know this, but to anyone who reads this can you suggest a way i can improve my math skills to a level where i could be able to solve questions like this easily