Find the maximum area of a trapezoid inscribed in a semicircle
1 answer:
Without loss of generality, we can assume the semicircle has a radius of 1 and is described by
y = √(1 - x²)
Then the shorter base has length 2x and the longer base has length 2. The area of the trapezoid is
A = (1/2)(2x+2)√(1-x²) = (1+x)√(1-x²)
Differentiating with respect to x, we have
A' = √(1-x²) + (1+x)(-2x)/(2√(1-x²)
Setting this to zero, we get
0 = (1-x²) +(1+x)(-x)
0 = 2x² +x -1
(2x-1)(x+1) = 0
x = {-1, 1/2} . . . . . -1 is an extraneous solution that gives minimum area
So, for x = 1/2, the area is
A = (1 + 1/2)√(1 - (1/2)² = (3/2)√(3/4)
A = (3/4)√3
Of course, if the radius of the semicircle is "r", the maximum area is
A = (r²·3·√3)/4
y = √(1 - x²)
Then the shorter base has length 2x and the longer base has length 2. The area of the trapezoid is
A = (1/2)(2x+2)√(1-x²) = (1+x)√(1-x²)
Differentiating with respect to x, we have
A' = √(1-x²) + (1+x)(-2x)/(2√(1-x²)
Setting this to zero, we get
0 = (1-x²) +(1+x)(-x)
0 = 2x² +x -1
(2x-1)(x+1) = 0
x = {-1, 1/2} . . . . . -1 is an extraneous solution that gives minimum area
So, for x = 1/2, the area is
A = (1 + 1/2)√(1 - (1/2)² = (3/2)√(3/4)
A = (3/4)√3
Of course, if the radius of the semicircle is "r", the maximum area is
A = (r²·3·√3)/4
You might be interested in
The
formula , where r! is 1*2*3*...r
is the formula which gives us the total number of ways of forming groups of r objects, out of n objects.
for example, given 10 objects, there are C(10,6) ways of forming groups of 6, out of the 10 objects.
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The bucket contains 6 orange balls and 7 yellow balls, that is a total of 13 balls.
The total number of groups of 5 balls out of the 13, that we can form is :
<span>1) What is the probability that, of the 5 balls selected at random, at least one is orange and at least one is yellow?
the event can happen as follows:
{1 orange, 4 yellow} or </span>{2 orange, 3 yellow} or {3 orange, 2 yellow} or {4 orange, 1 yellow}
that is, the only cases we are not considering are {5 orange} or {5 yellow}
5 orange balls can be selected in C(6, 5) = 6!/(5!1!)=6 different ways,
and 5 yellow balls can be selected in C(7, 5)=7!/(5!2!)=(7*6)/2=21 different ways.
so 5 orange balls or 5 yellow balls can be selected in 6+21=27 different ways,
this means that P(5 orange balls or 5 yellow balls)=27/1287=0.02
P({1 orange, 4 yellow} or {2 orange, 3 yellow} or {3 orange, 2 yellow} or {4 orange, 1 yellow}) = 1-P(5 orange balls or 5 yellow balls)=1-0.02=0.98
Part 2
The event "at least two are orange and at least two are yellow " can happen in the following ways:
{2 orange, 3 yellow} or {3 orange, 2 yellow}
2 orange, 3 yellow balls can be selected in C(6,2)*C(7,3) because
any selected 2 of the 6 orange balls, can be combined with any group of 3, out of the 7 yellow balls.
so we calculate:
thus,
P ({2 orange, 3 yellow} or {3 orange, 2 yellow})=
P (2 orange, 3 yellow) + P (3 orange, 2 yellow)=