0
$\begingroup$

I have another problem that I can't figure out.

A triangle har 2 of its summits in points (7.788,0,1.95), (0,7.788,1.95), and the last one on the curve with all the points (7.788,7.788,a^2+1.95), a is a real number. Calculate the area f(a) of the triangle as a function of a and calculate where it takes its minimal value.

I tried the following. Because I can't calculate vectors from points, I counted by hand:

u1 = {-7.788, 7.788, 0}
u2 = {0, 7.888, a^2}

Then I did that:

(Norm[Cross[u1, u2]])/2

The output was:

1/2 Sqrt[3773.86 + 60.6529 Abs[a]^4 + Abs[0. + 7.788 a^2]^2]

Which is not the right solution.

How should I do this?

$\endgroup$
12
  • 2
    $\begingroup$ How do you know it is not correct? $\endgroup$
    – Kuba
    May 9, 2018 at 9:14
  • 1
    $\begingroup$ Let's define A = {7.788,0,1.95}, B= {0,7.788,1.95} and F[a_]:={7.788,7.788,a^2+1.95}. You can Cross two edge vectors of the triangle emanating from a common corner like this: Norm[Cross[B-A,F[a]-A]]/2. The result is the same as yours. For minimization, you can use the observation that a is a minimizer of area if and only if a is a minimizer of Cross[B - A, F[a] - A].Cross[B - A, F[a] - A] (that's the square of the norm!). I think you can do it on your own from this point on (you know how to minimize a quadratic function, right?) $\endgroup$ May 9, 2018 at 9:20
  • $\begingroup$ @Kuba: because I entered it in the system and it was wrong. $\endgroup$
    – Dovendyr
    May 9, 2018 at 9:43
  • $\begingroup$ @HenrikSchumacher: no sorry I don't. When we had it in class I used to derivate and search for f'(a)=0 $\endgroup$
    – Dovendyr
    May 9, 2018 at 9:44
  • 1
    $\begingroup$ @HenrikSchumacher we can use ComplexExpand earlier which will help too $\endgroup$
    – Kuba
    May 9, 2018 at 10:26

2 Answers 2

1
$\begingroup$

You can also use Area:

Area[
 Triangle[{{7.788, 0, 1.95}, {0, 7.788, 1.95}, {7.788, 7.788, 
    a^2 + 1.95}}]]

1/2 Sqrt[3678.78 + 121.306 a^4]

Minimize[%, a]

{30.3265, {a -> -1.49931*10^-16}}

$\endgroup$
2
  • $\begingroup$ Nice! I have test on monday, hope I will remember that :) $\endgroup$
    – Dovendyr
    May 9, 2018 at 13:45
  • $\begingroup$ Be aware, if you are supposed to be learning vector calculations, this might not be an acceptable test answer! $\endgroup$
    – george2079
    May 9, 2018 at 15:58
1
$\begingroup$

Just summing up the code:

A = {7.788, 0, 1.95};
B = {0, 7.788, 1.95};
F[a_] := {7.788, 7.788, a^2 + 1.95}

h[a_] := Cross[B - A, F[a] - A].Cross[B - A, F[a] - A]; 
f[a_]:= Sqrt[h[a]]/2;
Solve[h'[a] == 0, a]

{{a -> 0}, {a -> 0.}, {a -> 0.}}

So, $a = 0$ is the minimizer. This illustrates that:

Manipulate[
 Graphics3D[{
   Point[{A, B, F[a]}],
   Text[f[a], F[a] + {1, 1, 1}],
   Triangle[{A, B, F[a]}]
   },
  PlotRange -> {-20, 20}],
 {{a, 0}, -10, 10}
 ]

For the answer of your exercise: The exercise asks for minimal value of the function f. So that's

f[0]

30.3265

$\endgroup$
4
  • $\begingroup$ Thank you so much for your help today! It does ask me the area for f(a) and the minimum for a=-.... ? $\endgroup$
    – Dovendyr
    May 9, 2018 at 10:37
  • $\begingroup$ You're welcome. Well, I guess, it isn't a good idea to replace teachers by computers, is it? $a=0$ or $a = 0.$ should be accepted as answer, then. As far as I can tell, you did nothing wrong. It is a pity that young people's enthusiam for mathematics gets broken by such unmathematical obstacles. Take as today's lesson that everybody makes mistakes sometimes, be it students, teachers, or computer programmers. And simply move on. $\endgroup$ May 9, 2018 at 10:44
  • $\begingroup$ I redid the problem and it seems it was me who was wrong. I had a slightly different solution than yours, you had 3678.78, I had 3773.86. I guess one really needs to do the vectors properly and not calculate by hand. I wish you a great day! $\endgroup$
    – Dovendyr
    May 9, 2018 at 11:30
  • $\begingroup$ Thanks! Have a great day, too! $\endgroup$ May 9, 2018 at 11:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.