Question

Find an equation of the line through the point (3, 5) that cuts off the least area from the first quadrant?

Answer 1

The general equation of a straight line is $y=mx+b,$ where m is a slope, and b is the intercept on the y-axis.

Suppose the line passing through the points (3, 5) and (0, b).

Then we have:

$5=3m+b,$

$3m=5-b,$

$m=\frac{5-b}{3}.$

We have found the slope, so the equation of the curve is:

$y=\frac{5-b}{3}x+b.$

Now we can find the intercept on the x-axis (y=0):

$0=\frac{5-b}{3}x+b,$

$\frac{5-b}{3}x=-b,$

$x=\frac{-3b}{5-b}=\frac{3b}{b-5}.$

The area of the right triangle is $S=\frac{1}{2}ab$ , so we have

$S=\frac{1}{2}b\frac{3b}{b-5}=\frac{3b^2}{2(b-5)}.$

To find its minimum we have to find the derivative:

$S'=\frac{6b\cdot 2(b-5)-3b^2\cdot 2}{4(b-5)^2}=\frac{12b^2-60b-6b^2}{4(b-5)^2}=$

$=\frac{6b^2-60b}{4(b-5)^2}=\frac{6b(b-10)}{4(b-5)^2}.$

$S'=0$ when b=0 or b=10.

So b=10 produce the least area, and

$m=\frac{5-b}{3}=\frac{5-10}{3}=-\frac{5}{3}.$

The equation of the line through the point (3, 5) that cuts off the least area from the first quadrant is

$y=-\frac{5}{3}x+10.$

Answer: $y=-\frac{5}{3}x+10.$