Asked by cool
Let x and y be to positive numbers whose product is 500:
(a) Find the maximum sum of x and y
(b) Find the minimum sum of x and y:
2. A cylindrical container can hold a volume of 1 liter. Find the dimensions
of the container that minimizes the surface area.
3. What are the dimensions of the rectangle with largest area that can be
drawn inside a circle of radius R?
4. Find the maximal area of a right triangle with hypotenuse of length L:
(a) Find the maximum sum of x and y
(b) Find the minimum sum of x and y:
2. A cylindrical container can hold a volume of 1 liter. Find the dimensions
of the container that minimizes the surface area.
3. What are the dimensions of the rectangle with largest area that can be
drawn inside a circle of radius R?
4. Find the maximal area of a right triangle with hypotenuse of length L:
Answers
Answered by
Steve
#1 I assume that x,y,z are integers, or there is no maximum sum. The minimum sum would be 3∛500 ≈ 23.811
500 = 2^2 5^3 so its factors are
1,2,250
1,4,125
1,5,100
1,10,50
1,20,25
2,2,125
2,5,50
2,10,25
5,5,20
5,10,10
5,25,4
Now pick the largest and smallest sums.
#2 v = πr^2h = 1
so, h = 1/(πr^2)
The surface area is
a = 2πr^2 + 2πrh
= 2πr^2 + 2πr/(πr^2)
= 2πr^2 + 2/r
da/dr = 4πr - 2/r^2
da/dr=0 when r=1/∛(2π)
so, h=∛(4/π)
#3 The diagonal of the rectangle is the diameter of the circle. So,
x^2+y^2 = 4r^2
a = xy = x√(4r^2-x^2)
da/dx = 2(2r^2-x^2)/√(4r^2-x^2)
da/dx=0 when x = r√2
So, as expected, the rectangle is a square with diagonal 2r
#4 as above, it will be half of a square, with legs L/√2.
It's really the same problem, because a right triangle can be inscribed in a circle. Its hypotenuse is the diameter of the circle.
500 = 2^2 5^3 so its factors are
1,2,250
1,4,125
1,5,100
1,10,50
1,20,25
2,2,125
2,5,50
2,10,25
5,5,20
5,10,10
5,25,4
Now pick the largest and smallest sums.
#2 v = πr^2h = 1
so, h = 1/(πr^2)
The surface area is
a = 2πr^2 + 2πrh
= 2πr^2 + 2πr/(πr^2)
= 2πr^2 + 2/r
da/dr = 4πr - 2/r^2
da/dr=0 when r=1/∛(2π)
so, h=∛(4/π)
#3 The diagonal of the rectangle is the diameter of the circle. So,
x^2+y^2 = 4r^2
a = xy = x√(4r^2-x^2)
da/dx = 2(2r^2-x^2)/√(4r^2-x^2)
da/dx=0 when x = r√2
So, as expected, the rectangle is a square with diagonal 2r
#4 as above, it will be half of a square, with legs L/√2.
It's really the same problem, because a right triangle can be inscribed in a circle. Its hypotenuse is the diameter of the circle.
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