For this case the first thing you should do is observe that the diameter of the four semicircles is the same.
Therefore, we can decompose the figure as follows:
1) We draw the diameters of the four semicircles to form a square.
2) We divide the figure into a square and four semicircles
3) The total area is the sum of the area of the square, plus the area of the 4 semicircles.
Answer:
c)as a square and four semicircles
57-36=21 times (rolled less than 3)
36:21 →12:7
Let the lengths of the east and west sides be x and the lengths of the north and south sides be y. the dimensions you want are therefore x and y.
The cost of the east and west fencing is $4*2*x; the cost of the north and south fencing is $2*2*y. We have to put in that "2" because there are 2 sides that run from east to west and 2 sides that run from north to south.
The total cost of all this fencing is $4(2)(x) + $2(2)(y) = $128. Let's reduce this by dividing all three terms by 4: 2x + y = 32.
Now we are to maximize the area of the vegetable patch, subject to the constraint that 2x + y = 32. The formula for area is A = L * W. Solving 2x + y = 32 for y, we get y = -2x + 32.
We can now eliminate y. The area of the patch is (x)(-2x+32) = A. We want to maximize A.
If you're in algebra, find the x-coordinate of the vertex of this quadratic equation. Remember the formula x = -b/(2a)? Once you have calculated this x, subst. your value into the formula for y: y= -2x + 32.
Now multiply together your x and y values to obtain the max area of the patch.
If you're in calculus, differentiate A = x(-2x+32) with respect to x and set the derivative equal to zero. This approach should give you the same x value as before; the corresponding y value will be the same; y=-2x+32.
Multiply x and y together. That'll give you the maximum possible area of the garden patch.
