### Video Transcript

A rectangular-shaped playground ends in two semicircles. Given that the perimeter of the playground is 594 meters, determine its maximum area.

Weβre asked to find the maximum area of a playground composed of a rectangle ending in two semicircles with a given perimeter of 594 meters. Letβs begin by labeling the width of the inner rectangle π₯, which is also the diameter of the semicircles, and letβs label the length of the rectangle π¦. Since there are two semicircles, one on each end of the rectangle, the area and perimeter of the two ends combined are that of a whole circle. And this circle has a radius of π, which is equal to π₯ over two.

Now recalling that the area of any circle is ππ squared, where π is the radius, we have the area of our circle equal to π multiplied by π₯ over two squared, that is, π times π₯ squared over four. Now recalling also that the perimeter of a circle is given by π is two ππ, we have the perimeter of our circle equal to two π multiplied by π₯ over two. And dividing numerator and denominator by two, the perimeter of our circle is π multiplied by π₯. And so, our two ends combine to make a circle with an area of π times π₯ squared over four and a perimeter of π times π₯.

Now, the area of our playground π΄ is the sum of the area of the inner rectangle and the total area of the circle. The area of the inner rectangle is π₯ multiplied by π¦. And weβve just found the area of the circle; thatβs ππ₯ squared over four. The perimeter is the length of the outline of the playground, that is, two times the length of the inner rectangle and the perimeter of our two semicircles, which we worked out is π multiplied by π₯.

Now, since weβre given that the perimeter of the playground is actually 594 meters, the optimization problem we want to solve is then maximize the area, thatβs π₯π¦ plus ππ₯ squared over four, given the constraint that the perimeter is two π¦ plus ππ₯, and thatβs equal to 594. Now, in order to maximize the area, weβre going to use the second derivative test. And in order to do this, we want to rewrite the total area of the playground as a function of π₯ only. And to do this, we make π¦ the subject of the constraint. Subtracting π times π₯ from both sides and dividing both sides by two, we have π¦ equal to 297 minus ππ₯ over two. And making some space, we want to substitute this into the area π΄. This gives us π΄ as a function of π₯ is equal to π₯ multiplied by 297 minus ππ₯ over two plus ππ₯ squared over four.

Now distributing our parentheses, this gives us 297π₯ minus ππ₯ squared over two plus ππ₯ squared over four. And this simplifies to 297π₯ minus ππ₯ squared over four. So now making some space, to maximize our area, we need to find the critical points. And to do this, we take the first derivative of our area as a function of π₯ and set it equal to zero and solve for π₯. Differentiating 297π₯ with respect to π₯ gives us 297.

And differentiating ππ₯ squared over four with respect to π₯, we multiply by the power, thatβs two, and take one from the power, which leaves one. π₯ to the power one is equal to π₯. And we can divide top and bottom by two so that the first derivative of the area function is 297 minus ππ₯ over two. And we need to set this equal to zero and so for π₯. We do this by adding ππ₯ over two to both sides. And multiplying both sides by two and dividing by π, we have two times 297 over π is equal to π₯. That is, π₯ is equal to 594 over π.

Now making some space, next, we need to determine the nature of this critical point. And we use the second derivative test to do this. And this tells us that given a twice differentiable function π with a stationary point at π₯ is equal to π₯ sub zero, if the second derivative π double prime of π is greater than zero at π₯ is equal to π₯ sub zero, then π₯ sub zero is a local minimum. If π double prime at π₯ sub zero is less than zero, then π₯ sub zero is a local maximum. And if π double prime at π₯ sub zero is equal to zero, then the test is inconclusive.

So now taking the second derivative of our function π΄, differentiating the constant 297 with respect to π₯ gives us zero, and differentiating negative π over two π₯ with respect to π₯ gives us negative π over two. Our second derivative is therefore the negative constant negative π over two for all values of π₯. And so, the second derivative at our critical point π₯ is 594 over π is negative π over two, which is less than zero. We see then that by the second derivative test our critical point is a local maximum. That is, π₯ is equal to 594 over π is a local maximum of the area function π΄ of π₯ of the playground.

So now making some space, if we substitute our local maximum π₯ is equal to 594 over π into our function π΄ of π₯, we can find the value of the maximum area. This gives us 297 multiplied by 594 over π minus π over four multiplied by 594 over π all squared. This evaluates to 176418 over π minus 352836 over four π, which in turn evaluates to 88209 over π.

The maximum area of the playground with perimeter 594 meters is therefore 88209 over π meters squared.