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Mathematics 14 Online
OpenStudy (anonymous):

the sum of the first number and 3 times the second number is 42. What are the numbers?

OpenStudy (agreene):

Do these numbers have to be in order? because the way this is written... there are infinite answers to this question.

OpenStudy (anonymous):

Oh, and they need to be 2 positive real numbers whose product is a maximum.

OpenStudy (agreene):

product is a maximum of what?

OpenStudy (anonymous):

Just the largest possible number.

OpenStudy (agreene):

x+3y=42 3y=42-x \[y=14-\frac x3\] You can use that equation to put in a number for x (first number) and it will give you y (the second number). The largest possible number will be whatever number you can think of that is really big, there is no real answer... In fact, there are still infinite answers to this question.

OpenStudy (anonymous):

What about graphing a parabola and using the vertex? That's what my teacher used on the example.

OpenStudy (agreene):

Hrm, yeah, I didnt think about this all the way through... since they have to be positive, and y = 14-(1/3)x is a downward sloping line, the y intercept will be the largest single number. so, first number 0 second number 14

OpenStudy (anonymous):

Thanks!

OpenStudy (anonymous):

Wait- is that the maximum product? 0?

OpenStudy (agreene):

Oh, you want a product that is a max, well thats going to be rather annoying since it's a line, lol... But you can find it using the distance formula and the points for the x and y intercepts. (0,14) (42,0) Remember: \[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\] \[d=\sqrt{(0-42)^2+(14-0)^2}\] \[d=\sqrt{1764+196}\] \[d=\sqrt{1960}\] \[d=14\sqrt{10}\] Now, thats telling us that the length of the hypotenuse of the triangle the line and the axes makes. 1/2 of this will give us the hypotenuse we need to determine the maximization for (x,y)

OpenStudy (anonymous):

Thank you!!!

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