<p>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, 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 200?</p>
<p>This is a Grid-In question and is 4100. Could someone explain? Thank you :)</p>
<p>Bump… Help?</p>
<p>You sure its 4100 do you mean year 2000 or year 200 ? if you mistyped year as year 200 then answer should be 500;</p>
<p>v = q + (r - 6y - 11820)^2;</p>
<p>500 = q + (r - 120)^2;
500 = q + r^2 - 240r + 14400;
-13900 + 240r = q + r^2;</p>
<p>v = q + (r - 6y - 11820)^2;
v = q + (r - 12000 - 11820)^2;
v = q + (r - 180)^2;
v = q + r^2 -360r + 32400;</p>
<p>We know what q + r^2 = -13900 + 240r;</p>
<p>v = -13900 + 240r - 360r + 32400
v = 18500 -120r;</p>
<p>500 = 18500 - 120r;
500 - 18500 = -120r;
-18000 = -120r;
r = 150;</p>
<p>
Sorry, it was 2000</p>
<p>
It was the answer on the answer key on one of the PR test…</p>
<p>Here is what is says:
18. 4100
The minimum value of any real number squared is 0, since 0 = 0^2 and anything else squared is positive. So the minimum value of the expression is parentheses is 0. The lowest value (v) of the painting must therefore be equal to q and, since it occurred in 1990, (r - 6(1990 - 1970))^2 = 0. Therefore, (r - 6(1990 - 1970)) = 0. Solving this equation for r yields r = 120. Since the dollar value of the painting in 1990 is 500, q = 500. Now plug in 2000 for the year, and solve for v: v = 500 + (120 -6(2000-1970))^2 = 4100.</p>
<p>… I really don’t get what they are talking about…</p>
<p>Okay I solved it you provided me by certain hint:
You have to remember though the minium of any real number squared isn’t 0
2^0 = 1;</p>
<p>Here your answer Plus explanation:-</p>
<p>v = q + (r - 6(y - 1970))^2;</p>
<p>v = q + (r - 6y - 11820)^2;</p>
<p>Lets compute for 1990:-</p>
<p>500 = q + (r - 6(1990) - 11820)^2;
500 = q + (r - 11940 - 11820)^2;
In order for that equation to hold true we know that the numbers inside the parenthesis squared would be = 0 because they told us the miniumum,so based on that info we know that:
q = 500;</p>
<p>So lets solve for R:</p>
<p>500 = 500 + (r - 120)^2;</p>
<p>In order to make (r - 120)^2 = 0 we have to make r = 120 because
120 - 120 = 0^2 = 0; So</p>
<p>r = 120;</p>
<p>Now lets solve for year 2000
v = 500 + (r - 6(2000) - 11820)^2;
v = 500 + (120 - 180)^2
v = 500 + (-60)^2;
v = 500 + 3600 = 4100;</p>
<p>This required little logic plus some algebra to solve it I wonder what be the complete algebraic solving of this would be.</p>
<p>I hope DrSteve or xigi sees this topic because I am really interesting to see just the algebraic way of solving this.</p>
<p>I don’t have time to read through the other solutions right now, so I apologize if I’m repeating a solution already posted. I’m just going to make the outermost parentheses () into brackets [] for ease of reading:</p>
<p>v = q + [r - 6(y - 1970)]^2</p>
<p>The minimum occurs when r - 6(1990 - 1970) = 0, or r - 120 = 0, or r = 120.
So the equation becomes:</p>
<p>v = q + [120 - 6(y - 1970)]^2</p>
<p>Now using the fact that v=500 when y = 1900 we have:</p>
<p>500 = q + [120 - 6(1990 - 1970)]^2 = q + [120 - 6(20)]^2 = q + (120- 120)^2 = q. So q = 500 and the equation becomes:</p>
<p>v = 500 + [120 - 6(y - 1970)]^2</p>
<p>Now plug in 2000 for y.</p>
<p>v = 500 + [120 - 6(2000 - 1970)]^2 = 500 + [120 - 6(30)]^2 = 500 + (-60)^2 = 500 + 3600 = 4100. </p>
<p>This is a very difficult problem (relatively speaking for SAT math). Let me know if you need me to clarify any of the steps.</p>
<p>Question for the OP: where did you get this problem?</p>
<p>@DrSteve</p>
<p>I wanted to ask you is it possible to get r,v,q without coming with conclusion that
(r - 6(1990) - 11820)^2 = 0 ? By doing pure algebra without coming with this logical conclusion ?</p>
<p>If he told us that there isn’t minimum value how would we solve this algebraically?</p>
<p>@Generic:</p>
<p>There are 2 unknown constants in this question: r and q. Therefore we probably need 2 pieces of information to find both constants. In this case we are given the point (1990, 500), AND we are given the minimum value occurs when y = 1990. I don’t see a way to avoid using one of these pieces of information.</p>
<p>@DrSteve</p>
<p>I am just out of curiosity plus if one question and if question doesn’t mention that its the minimum value how we would be solving for both r and q.</p>
<p>But v of 2000 isn’t not given I tried to solve this first using pure algebra and got r^2 + q in terms of year 1990 not year 2000,so I got bad result as you can see in my first post.</p>
<p>^Just wanted to thank you both :)</p>