Using Parabolas to Find Numbers so sum is 50, sum of sqrs is

[muk]

Use parabolas to work the following problem: find two numbers having a sum of 50 such that the sum of the square of one and three times the square of the other is a minimum.

To be honest, I'm really not sure how to start. I understand that a minimum is the lowest point on a graph, and that in the context of a positive parabola it is the vertex. I haven't the foggiest idea how a sum of squares can be a minimum, because as far as I know, a minimum is an ordered pair, not a number. So far, I have:

x[sub:33bt8t68]1[/sub:33bt8t68] + x[sub:33bt8t68]2[/sub:33bt8t68] =50
x[sub:33bt8t68]1[/sub:33bt8t68][sup:33bt8t68]2[/sup:33bt8t68] + 3x[sub:33bt8t68]2[/sub:33bt8t68][sup:33bt8t68]2[/sup:33bt8t68] = minimum

 

[galactus]

Re: Using Parabolas to Find Numbers

You are on the right track.

\(\displaystyle x+y=50\)..........[1]

\(\displaystyle x^{2}+3y^{2}=S\)..........minimize this.

Solve [1] for x or y and sub into S. Then, differentiate to get \(\displaystyle \frac{dS}{dx} \;\ or \;\ \frac{dS}{dy}\)(depending on which variable you solve for), set to 0, and solve.

You can also do this without calc by noting that, aftere you sub, the function is a quadratic.

 

[muk]

Re: Using Parabolas to Find Numbers

Thanks very much, but I'm still having trouble solving the problem.

I am trying to do it with quadratics, assuming that I won't have to find derivatives in a precalculus course where I haven't been taught to use derivatives.

I start by solving equation one for x

x + y = 50
x = y - 50

Then I substitute y - 50 for x in equation two.

x[sup:3d3gs6ke]2[/sup:3d3gs6ke] + 3y[sup:3d3gs6ke]2[/sup:3d3gs6ke] = S

(y - 50)[sup:3d3gs6ke]2[/sup:3d3gs6ke] + 3y[sup:3d3gs6ke]2[/sup:3d3gs6ke] = S

y[sup:3d3gs6ke]2[/sup:3d3gs6ke] - 100y + 2500 +3y[sup:3d3gs6ke]2[/sup:3d3gs6ke] = S

4y[sup:3d3gs6ke]2[/sup:3d3gs6ke] -100y + 2500 = S

I assume that I take S to equal zero. I'm not sure about this , but I don't know what else I would do.

4y[sup:3d3gs6ke]2[/sup:3d3gs6ke] -100y + 2500 = 0

This is not a product of two square binomials, so I divide B by 2 and square it to get 625, then I subtract 1875 from both sides.

4y[sup:3d3gs6ke]2[/sup:3d3gs6ke] -100y + 625 = -1875

(2y-25)[sup:3d3gs6ke]2[/sup:3d3gs6ke] = -1875

Take the square root of both sides.

2y-25 = 25i*sqrt3
y = (25+25i*sqrt3)/2

When I substitute this into equation 1, I get x = (75-25i*sqrt3)/2

However, when I substitute both x and y into equation 2, I do not get 0. I also get a different x by going back to the beginning and solving equation 1 for y instead of x.

Sorry for the trouble.

 

[Deleted member 4993]

Re: Using Parabolas to Find Numbers

Thanks very much, but I'm still having trouble solving the problem.

I am trying to do it with quadratics, assuming that I won't have to find derivatives in a precalculus course where I haven't been taught to use derivatives.

I start by solving equation one for x

x + y = 50
x = y - 50

Then I substitute y - 50 for x in equation two.

x[sup:3sxe9h67]2[/sup:3sxe9h67] + 3y[sup:3sxe9h67]2[/sup:3sxe9h67] = S

(y - 50)[sup:3sxe9h67]2[/sup:3sxe9h67] + 3y[sup:3sxe9h67]2[/sup:3sxe9h67] = S

y[sup:3sxe9h67]2[/sup:3sxe9h67] - 100y + 2500 +3y[sup:3sxe9h67]2[/sup:3sxe9h67] = S

4y[sup:3sxe9h67]2[/sup:3sxe9h67] -100y + 2500 = S

This describes a parabola where

S = (2y - 25)[sup:3sxe9h67]2[/sup:3sxe9h67] + 1875

S is minimum when

2y - 25 = 0

y = 12.5

Now continue....

I assume that I take S to equal zero. I'm not sure about this , but I don't know what else I would do.

4y[sup:3sxe9h67]2[/sup:3sxe9h67] -100y + 2500 = 0

This is not a product of two square binomials, so I divide B by 2 and square it to get 625, then I subtract 1875 from both sides.

4y[sup:3sxe9h67]2[/sup:3sxe9h67] -100y + 625 = -1875

(2y-25)[sup:3sxe9h67]2[/sup:3sxe9h67] = -1875

Take the square root of both sides.

2y-25 = 25i*sqrt3
y = (25+25i*sqrt3)/2

When I substitute this into equation 1, I get x = (75-25i*sqrt3)/2

However, when I substitute both x and y into equation 2, I do not get 0. I also get a different x by going back to the beginning and solving equation 1 for y instead of x.

Sorry for the trouble.

 

[soroban]

Re: Using Parabolas to Find Numbers

Hello, muk!

Use parabolas to solve: .Find two numbers having a sum of 50 such that
the sum of the square of one and three times the square of the other is a minimum.

You started correctly . . .

\(\displaystyle \text{Let the two numbers be }x\text{ and } y.\)

\(\displaystyle \text{We have: }\;\begin{array}{cccc}x + y &=& 50 &[1] \\ x^2+3y^2 &=&f & [2]\end{array}\)

\(\displaystyle \text{From [1] we have: }\;x \:=\:50-y\;\;[3]\)

\(\displaystyle \text{Substitute into [2]: }\;f \;=\;(50-y)^2 + 3y^2 \quad\Rightarrow\quad f \;=\;4y^2 - 100y + 2500\)


This is an up-opening parabola; its minimum value is at its vertex.

. . \(\displaystyle v \:=\:\frac{\text{-}b}{2a} \quad\Rightarrow\quad y \:=\:\frac{\text{-}(\text{-}100)}{2(4)} \quad\Rightarrow\quad\boxed{ y\:=\:\frac{25}{2}}\)

\(\displaystyle \text{Substitute into [3]: }\;x \;=\;50-\frac{25}{2} \quad\Rightarrow\quad\boxed{ x\;=\;\frac{75}{2}}\)

 

[galactus]

Re: Using Parabolas to Find Numbers

If you are not in calculus, then Soroban's way is the best. If you want to see the calculus method, just differentiate

\(\displaystyle S=4y^{2}-100y+2500\)

\(\displaystyle \frac{dS}{dy}=8y-100\)

Set to 0 and solve for y:

\(\displaystyle 8y-100=0\)

Solving, we easily see that \(\displaystyle y=\frac{25}{2}\)

Of course, since they sum to 50, we see that \(\displaystyle x=\frac{75}{2}\)

Sub them back in to see if they work...........they do.

 

[muk]

Re: Using Parabolas to Find Numbers

Ok, thanks guys! It was an awful lot to put together, but I get it now.

Thanks a lot for all of the help!