Determine the equation of the line through the point (3,5) that cuts off the least area of the first quadrant.
By using Point Slope Form, we can determine the equation of the line that pass through the point (3,5).
y−5=m(x−3)y=mx−3m+5
So the y-intercept when x= will be,
y=m(0)−3m+5
And the x-intercept when y=0 will be,
0=mx−3m+5x=3−5m
So, the area of the triangle will be A=12(3−5m)(5−3m)A=15−252m−9m2
If we take the derivative of A, we get...
A′=252m2−92
when A′=0,
252m2=92m2=259m=−53 and m=53
m=−53 respects a minimum since A″
\begin{equation} \begin{aligned} \text{Therefore, the equation of the line is } y &= mx - 3m + 5 = - \frac{5}{3} x - 3 \left( \frac{-5}{3}\right)+5\\ \\ y &= \frac{-5}{3}x + 10 \end{aligned} \end{equation}
No comments:
Post a Comment