<p>a line crosses both the x and y-axis at a positive value (so the line has a negative slope) and contains the point (1,2)</p>
<p>find the y-coordinate that makes the triangle have a minimum area</p>
<p>i had something like:
A=1/2xy
5=rad(x^2+y^2)
25=x^2+y^2
y^2=25-x^2
y=rad(25-x^2) (not negative, obviously bc y can't be negative)</p>
<p>so A=1/2x<em>rad(25-x^2)
A=[1/2</em>rad(25-x^2)]+[1/2x<em>1/2</em>-2x*(25-x^2)^-1/2]
A=rad(25-x^2)/2 - x^2/2rad(25-x^2)
A=25-x^2-x^2/2rad(25-x^2)
A=25-2x^2/2rad(25-x^2)
When A'=0, x=-sqrt12.5, +sqrt12.5
and when you plug in x=sqrt12.5, y=sqrt12.5 as well</p>
<p>but A'>0 before x=sqrt12.5, A'=0 at sqrt12.5, and A'<0 after sqrt12.5, making it a maximum</p>
<p>i somehow bs'd something with the 2nd derivative test and made it a minimum, but is my answer even right?</p>
<p>i got x=sqrt(12.5) and y=sqrt(12.5), which turned out to be (0, 3.5355 ish) but that turned out to be the maximum</p>
<p>a line crosses both the x and y-axis at a positive value (so the line has a negative slope) and contains the point (1,2)</p>
<p>find the y-coordinate that makes the triangle have a minimum area</p>
<p>i had something like:
A=1/2xy
5=rad(x^2+y^2) <-------- if i didn’t use the 5, then all i could get was perimeter=x+y+rad(x^2+y^2), so i just used 5 bc 2^2+1^2=5, but i guess it’s probably wrong
25=x^2+y^2
y^2=25-x^2
y=rad(25-x^2) (not negative, obviously bc y can’t be negative)</p>
<p>so A=1/2x<em>rad(25-x^2)
A=[1/2</em>rad(25-x^2)]+[1/2x<em>1/2</em>-2x*(25-x^2)^-1/2]
A=rad(25-x^2)/2 - x^2/2rad(25-x^2)
A=25-x^2-x^2/2rad(25-x^2)
A=25-2x^2/2rad(25-x^2)
When A’=0, x=-sqrt12.5, +sqrt12.5
and when you plug in x=sqrt12.5, y=sqrt12.5 as well</p>
<p>but A’>0 before x=sqrt12.5, A’=0 at sqrt12.5, and A’<0 after sqrt12.5, making it a maximum</p>
<p>i somehow bs’d something with the 2nd derivative test and made it a minimum, but is my answer even right?</p>
<p>i got x=sqrt(12.5) and y=sqrt(12.5), which turned out to be (0, 3.5355 ish) but that turned out to be the maximum</p>
<p>If x = t is the x-intercept, then
y = -2x/(t-1) + 2t/(t-1), t>1.
the y-intercept is 2t/(t-1).
A = (1/2)(x-intercept)(y-intercept)
A = t^2 / (t-1)
A’ = (t^2 - 2t) / (t-1)^2.
A’=0 when t=2.
You can prove that A(2) is the minimum.
y = 2t/(t-1) for t=2
y = 4.</p>