Tutorial Exercise

Find the equation of the line through the point $(4,7)$ that cuts off the least area from the first quadrant.

Step $1$

A line with slope $m$ through $(4,7)$ has equation $y=mx+7-4m$

$.($Hint: Solve for b$.)$

Step $2$

The $y-$intercept is $7-4m$ and the $x-$intercept is

m

Step $3$

Therefore, the area to be minimized is given by

$A(m)=\frac{1}{2}(7-4m)(\frac{-7}{m}+4)$

$=28-\frac{494}{2m}-8.$

Step $4$

$49$

Now, $A^{\text{'}}(m)=$

${?}_{2m^2}-8$

Step $5$

Since $m$ is negative, then solving $A^{\text{'}}=0$ gives $m=-\frac{7vv,7}{4}.$

Step $6$

So the equation of the line which minimizes the area is