A triangle has two of its corners in (3.052 , 0 , 1.456), (0 , 3.052, 1.456), and the third of the curve in the room consisting of all points (3.052 , 3.052, a ^ 2 + 1.456), where a is a real number. Calculate the area f(a) of the triangle and state where the function assumes its minimum value. (Positive oriented ON system.) The area can be expressed as a function.
How i tried using mathematica.
u1 = {3.052, 0, 1.456}
u2 = {0, 3.052, 1.456}
u3 = {3.052, 3.052, a^2 + 1.456}
v1 = u2 - u1
v2 = u3 - u1
Norm[Cross[v1, v2]/2]
Which gives the area: Sqrt[21.690927651904005+ 1/2 Abs[0. + 3.052` a^2]^2]
And using the Minimize function: Minimize[Sqrt[21.690927651904005+ 1/2 Abs[0. + 3.052` a^2]^2], a]
gives the minimum value for a = 5.87669*10^-7
This is wrong or is in the wrong form, would appreciate any help!
Let 3.052=p and 1.456=q, so the points are (say) A$(p,0,q)$; B$(0,p,q)$ and C$(p,p,a^2+q)$
Now, $\vec {AB}=(-p,p,0)$ and $\vec {BC}=(p,0,a^2)$
$$\vec {AB} \times \vec {BC}=(a^2p,a^2p,-p^2)$$
Area=$\frac{1}{2}|\vec {AB} \times \vec {BC}|=\frac{1}{2} \sqrt{2a^4p^2+p^4}=\frac{1}{2}p \sqrt{2a^4+p^2}$
$\frac{dA}{da}=\frac{2pa^3}{\sqrt{2a^4+p^2}}=0 \Rightarrow a=0$