Difficult Word Problem with multiple variable

[vortex705]

The question is:

In year y, the value in dollars, v, of a certain painting created in 1970 is given by the equation v = q + (r - 6( y- 1970))^2[/color], where q and r are constants. If the painting reached its lowest value in 1990, when it was worth $500, what was the painting's value, in dollars, in the year 2000?

If it helps the answer is: The value of painting is $4100 in year 2000

Can anyone show me how I can approach this problem in order to get the answer? Thank you.

 

[fasteddie65]

The question is:

In year y, the value in dollars, v, of a certain painting created in 1970 is given by the equation v = q + (r - 6( y- 1970))^2[/color], where q and r are constants. If the painting reached its lowest value in 1990, when it was worth $500, what was the painting's value, in dollars, in the year 2000?

If it helps the answer is: The value of painting is $4100 in year 2000

Can anyone show me how I can approach this problem in order to get the answer? Thank you.

v = q + (r - 6(2000 - 1970)^2

v(1990) = q + r(1990 - 1970)^2 = 500
500 = q + r(20^2) = q + 400r

The difficulty is that not enough information is given to find the values of q and r. We have two variables, so to find a unique solution we need two equations.

 

[galactus]

You need to find q and r. We are told that the paintings lowest price is when y=20 (1990-1970). You can start by using the derivative to find that low point.

We know the low point is when y=20 and that v=500 at that point. You can use this to find a variable.

 

[vortex705]

I realized that the graphical representation would be a parabolic, and that minimum being the year = 1990 and $500.

Can you please show me the step to arrive at the answer. And if there is other way to solve this problem other than using derivative can you show me? The recommended time to solve this problem is under one minute.

 

[galactus]

You do not have to use the derivative. We can use the vertex of a parabola formula. Remember it?. \(\displaystyle x=\frac{-b}{2a}\)

If we expand out and group, we get:

\(\displaystyle v=\overbrace{36}^{\text{a}}y^{2}-\underbrace{(12r+141840)}_{\text{b}}y+\overbrace{r^{2}+23640r+q+139712400}^{\text{constant}}\)

By using the formula, we have \(\displaystyle \frac{12r+141840}{72}=20\)

Solve for r. Then, sub that back into the original along with v=500, y=20, then solve for q.

You'll have q and r found, so sub in y=30 and you got it. Let me know what you get for q and r. It works out, though this may not be the slickest way.