ok i'll explain the second one.
I normally don't do a whole question, but you asked for the solution so i'll show you and try to explain how to do these questions. **The diagram is important.
look at the diagram i have attached. You should label your diagram similarly and draw it up completely in the exam.
This is how you approach all of these questions, they're all very similar.
First look at what you're given, you have perimeter.
The idea is to create two equations, substitute into each of them, and differentiate to find a max/min.
Try to create an equation, using algebra, with as few variables as possible.
so we can see obviously that the shape gives us,
P = c + 2l + d
now c, (the circumference of half the circle) can be expressed using it's diameter. c = pi*d / 2 (divide by 2 because it's the arc of a semicircle)
therefore P = 2l + d + pi*d/2 {only 2 variables P=4}
4 = 2l + d*(1+pi/2)
Now we need to find a formular for Area. Since it's asking for a max you will be differentiating, so you'll want an equation.
A = d*l {for the rectangle} + pi*(d/2)^2 /2 {for the semicircle}
A = d*l + pi*d^2/8
So we will make the subsitution of l since it is easier to solve for and won't complicate very much. From P = 4
l = 2 - d*(1 + pi/2)/2
into A
A = d*[2 - d*(1+pi/2)/2)] + (pi/8)*d^2
= 2d - [1/2 + pi/4]*d^2 + (pi/8)*d^2
= 2d - (1/2 + pi/8)*d^2 - you can simplify this to [2d - anumber*d^2] if you want though i don't take credit for it.
so here is our quadratic for A with one variable d.
**Now you can cheat since it's a parabola and say that it's max/min must be halfway between it's solutions which you can easily find by factoring.
To find the max or min of anything, we differentiate it and make it equal zero. Diffing with respect to d
A' = 2 - (1/2 + pi/8)*2*d = 0
= 2 - (1 +pi/4)*d
= (1 + pi/4)*d= 2
d = 2/(1 + pi/4) = 1.12
So we know where our quadratic for the max or min of the area equals
Subbing this back into the equation for Area gives us
A = 1.12 this is exactly the same as d. So i suspect i made an error somewhere because i haven't done it on paper yet. I'll attach an edit when i do it on paper.
* You're obliged to, when doing max and min questions, to determine if you have calculated a max or a minimum. you can use a table of values, take the second derivative, or you can cheat and use the graph. Because it's a negative quadratic (the leading coefficient of d^2 is negative) the parabola will be upside down, hence it's a maxima.
If someone finds an error in my work please correct it.
