trig hw help?

Question

a maximum and minimum problems


1. find two numbers such that thier sum is 20 and the sum of thier squares is a minimum

2. find two numbers such that thier sum is 30 and the sum of thier squarres is a minimum

please explain how to do it

Answer 1

To do this you need to solve for one of the numbers, by setting up an equation.

x+y=20
y=20-x

Plug this into the other equation and simplify:
x^2+y^2=f
x^2+(20-x)^2=f
x^2+400-40x+x^2=f
2x^2-40x+400=f

Find the minimum either by graphing or the vertex formula (-b/(2a)) and plug it into y=20-x to find the other number.

The second question is done in the same way.

Answer 2

Ok, this is how you do it with calculus.

let's say a= first number b=second numbers

a + b = 20

a^2+b^2 = some number

now you solve the first equation for a or b, it doesn't matter

a = 20 - b

and you substitute

(20-b)^2+b^2= some number

now you opperate

(b^2-40*b+400)+b^2 = 2*b^2-40*b+400

Now you can do two things, you can graph or divide, if you graph, then graph the function f(b)=2*b^2-40*b+400 now if you derive you get

f'(b)= 4*b-40

and you make it equatl to 0

4*b-40 = 0

and you solve for b

b = 40/4 or b = 10

and now you can get a

a = 20-b

a = 20-10 = 10

so then you get that a = 10 and b = 10 that will give you the minimum you want and you do the same with number two

a + b = 30

a^2+b^2 = some number

now you solve the first equation for a or b, it doesn't matter

a = 30 - b

and you substitute

(30-b)^2+b^2= some number

now you opperate

(b^2-60*b+900)+b^2 = 2*b^2-60*b+900

Now you can do two things, you can graph or divide, if you graph, then graph the function f(b)=2*b^2-60*b+900 now if you derive you get

f'(b)= 4*b-60

and you make it equal to 0

4*b-60 = 0

and you solve for b

b = 60/4 or b = 15

and now you can get a

a = 30-b

a = 30-15 = 15

so then you get that a = 15 and b = 15 that will give you the minimum you want

hope this helps.

pd. equal to 0 when you derive because that means that the slope is going to be 0 and the slope becomes 0 when it's maximum or minimum values.

Answer 3

This is a problem that is easy to solve using calculus, but I don't know if that's the kind of help you want. You can take this or leave as you wish:

Two numbers that sum to 20 are x and (20 - x).

The squares of these numbers are x^2 and 400 - 40x + x^2.

The sum of the squares as a function of x is:
f(x) = 400 - 40x + 2x^2

To find extrema, we can take the derivative of f(x), set it equal to zero, and solve for x.

f'(x) = -40 + 4x = 0
4x = 40
x = 10
20 - x = 10.

One might now observe that to minimize the sum of the squares, the numbers should be equal.

So two numbers whose sum is 30 that have a minimal sum of squares are 15 and 15.

We can show this as above:

x and 30 - x sum to 30
x^2 and 900 - 60x + x^2 are the squares of the two numbers.
f(x) = 900 - 60x + 2x^2 is the sum of the squares as a function of x.

f'(x) = -60x + 4x = 0
4x = 60
x = 15
30 - x = 15

Normally, we'd use the second derivative test to figure out if the critical points of f'(x) are maxima or minima. But since f(x) is a parabola that points upward in both cases, the vertex of the parabola must be the minimum.

When I think maxima and minima, I think about calculus and not trig. But someone else out there might have a different suggestion.

EDIT:

I think of calculus so fast that the algebraic methods don't always register right away. With that said, I don't consider the algebraic methods for solving the problem to be trigonometric either, even those those topics (algebra and analytic geometry) are taught along with trigonometry in some classes.

Cheers, everyone.

Answer 4

x+y=20 y=20-x
S=x^2 + y^2= x^2+400-40x+x^2
S=2x^2-40x+400=2(x^2-20x)+400

this problem can be solved with calc or without

without calc, we can complete the square and find the vertex. the vertex of this concave up parabola will be the minimum

S=2(x^2-20x+100)+400-200
S=2(x-10)^2+200
the vertex is at x=10
if x=10 then y=20-x=10 so the 2 numbers are 10 and 10

with calculus
S'=4x-40 set S' equal to zero
x=10 when S'=0
this means at x=10, there is a min or a max
to see which one use S''
S''=4 which is positive and shows the graph is concave up
therefore x=10 must be a min and y=20-x=10
so the numbers are again 10 and 10

Answer 5

1a) x is 1st #
y is 2nd #
Therefore x + y = 20

and x^2 + y^2 = A --> x^2 + (20-x)^2 =A
Therefore:
2x^2-40x+400 =A
Take derivative:
A' = 4x -40
=4(x-10)
Therefore x = 10
And y = 10 (sub x into x+y=20)
Repeat the exact same thing for the second question

Answer 6

1. 10 and 10

2. 15 and 15

Answer 7

tan3x = tan(x + 2x) = (tanx + tan2x) / (a million - tanx tan2x) = [tanx + 2tanx/(a million - tan^2 x)] / [a million - tanx * 2tanx/(a million - tan^2 x)] = [(tanx - tan^3 x + 2tanx) / (a million - tan^2 x)] / (a million - tan^2 x - 2tan^2 x) / (a million - tan^2 x)] = (3tanx - tan^3 x) / (a million - 3tan^2 x)