1: If y = e2x, how would you show that y” + y' - 6y = 0. Explain the steps and show y

Step 1: Start with the given equation y = e^(2x).

1: If y = e2x, how would you show that y” + y' -...

1: If y = e2x, how would you show that y'' + y' - 6y = 0. Explain the steps and show you would prove that y'' + y' - 6y = 0 [5] 2: One side of a rectangle lies on the x-axis. The opposite side of the rectangle has vertices which lie on the graph of f(x) = -2e-0.5x2. For what value(s) of x is the area of the rectangle a maximum? What is the maximum area?

Transcript

00:01 In this question we need to inscribe a rectangle in the ellipse.

00:05 So the equation of ellipse is given to us which is x squared divided by 4 plus y square is equals to 1 and y is greater than 0.

00:17 And we also have to maximize the area of the rectangle.

00:22 So we will let the length of rectangle.

00:32 Be equals to l and width of rectangle be equals to w.

00:46 So area of the rectangle will be equal to l multiplied by w which is length multiplied by the width so here we put length as 2x multiplied by the width which is 2 y so it will be 4 xy and we have to maximize the area.

01:11 So we will make y as a subject from the equation of the ellipse.

01:19 So a which is the area will be equals to 4 x multiplied by if we will make y from the subject of the equation of the ellipse.

01:31 So it will be 1 minus x squared divided by 4 inside the root.

01:38 So if we will simplify it, we will get 2x square root of 4 minus x square.

01:48 So this is a function of the area.

01:50 So we will put a dash of x is equal to 0 for finding the critical points.

01:58 So on differentiating the function, we will get.

02:04 2 square root of 4 minus x squared plus 2x multiplied by minus 2x divided by square 2 2 minus x squared and 8 it will be 2 so it should be equals to 0 so from here on simplifying it we will get 8 minus 4 x square divided by square 2 root of 4 minus x square is must be equals to 0 so from here we will get the well value of x will be equals to square root of 2.

02:37 So now we will double differentiate it to find whether it is maximum or minimum...

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