<p>Can somebody please show me the steps to solving this problem? I keep getting the tangent line of f(x) to be y=12. I don't think that's right.</p>
<p>Consider f(x) = 12 - x^2 in [0, 2(sqrt3)]. Let A(t)be the area of the triangle formed by the coordinate axes and the tangent of the graph of f at the point (t, 12 - t^2). For what value of t is A(t) a minimum?</p>
<p>I don't wanna do it for you, but first find A(t). A(t) = 1/2(base)(height). Then find the minimum value of t for that equation.</p>
<p>Shouldn't I first find the tangent line equation?</p>
<p>Maybe this helps:</p>
<p>f(x) = 12-x^2
f'(x) = -2x</p>
<p>f'(t) = -2t</p>
<p>y-(12-t^2) = -2t(x-t)
y-12+t^2 = -2tx + 2t^2
y = t^2 - 2tx + 12 < Equation of tangent line at point t.</p>
<p>If y = t^2 - 2tx + 12 is the tangent line, then to find the base and the height you just have to find the x and y intercepts respectively.</p>
<p>Base.<br>
0 = t^2 -2tx + 12
x = (t^2 + 12)/2t</p>
<p>Height.
y = t^2 + 12</p>
<p>Area = .5 * base * height = (t^2 + 12)^2/4t
0 = 16t^2(t^2 + 12) - 4t(t^2 + 12) all that divided by 16t^2, but it won't matter
0 = 16t^4 -4t^3 + 192t^2 - 48t
0 = 4t(4t^3 - t^2 + 48t - 12)
t = 0 or t = 2.0000002</p>
<p>So, errr, two? Is that the answer?</p>
<p>The area function is correct, not sure what you did after that, I don't see what you are doing step by step. But if you didn't make any errors that should be right.</p>
<p>tell me what you mean by sqrt3 and I will help you</p>
<p>Anyways, if you're still there, you have to take the derivative of the equation of the area of the triangle. Then you apply the first derivative test. Gnight.</p>
<p>Thanks guys, I really appreciate the help. I think I finally understand the process required in working the problem. I'm gonna go watch Shrek 2, and after I do that I shall actually work the problem. Lol. Thanks again!</p>
<p>What I did was, I got the function (the are of a triangle) with respect to variable t. Then I found its critical points (more specifically its minima) by setting the first derivative equal to 0 or DNE.</p>