MHB Kylie's question at Yahoo Answers regarding minimizing the area of a triangle

[MarkFL]

Here is the question:

Kellie discovers a burn in her new scarf. She decides the best solution is to cut off one corner of the scarf?


The scarf is a rectangle and the burn is located 2cm from one edge and 5 cm from a perpendicular edge, what is the smallest area of the triangle she has to cut off to remove the burn?

using optimization!

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello kylie,

Let's orient our coordinate axes such that one corner of the scarf is at the origin and the burn is at the point $(2,5)$. The cut Kellie makes will lie along the line having $x$-intercept $(b,0)$ and $y$-intercept $(0,h)$, and must pass through the point at which the burn is located.

Using the two-intercept form for a line and the point at which the burn is located, we obtain the constraint:

$$\frac{2}{b}+\frac{5}{h}=1$$

Solving for $h$, we find:

$$h=\frac{5b}{b-2}$$

The area $A$ of the triangular portion of the scarf that is removed is:

$$A=\frac{1}{2}bh$$

Substituting for $h$, we obtain the area as a function of $b$ alone:

$$A(b)=\frac{b}{2}\left(\frac{5b}{b-2} \right)=\frac{5b^2}{2(b-2)}$$

Differentiating with respect to $b$ and equating the result to zero, we obtain:

$$A'(b)=\frac{2(b-2)(10b)-5b^2(2)}{(2(b-2))^2}=\frac{5b(b-4)}{2(b-2)^2}=0$$

Observing that we require $$2<b$$ we obtain the critical value:

$$b=4$$

Observing also that to the left of this critical value $A'(b)<0$ and to the right $A'(b)>0$, the first derivative tells us the critical value is at a minimum.

Hence:

$$A_{\min}=A(4)=20$$

Thus, the smallest triangular piece of scarf that Kellie can remove has an area of $20\text{ cm}^2$.