Science:Math Exam Resources/Courses/MATH100/December 2010/Question 05/Solution 1

According to the picture,

Thus, the cost equation is

$\displaystyle C=40d+20(8-x)$

To get this in terms of one variable, note by the Pythagorean theorem, we have

$\displaystyle d^{2}=6^{2}+x^{2}$

Hence, we have

$\displaystyle C=40{\sqrt {36+x^{2}}}+20(8-x)$

Differentiating yields

${\begin{aligned}C'&={\frac {40(2x)}{2{\sqrt {36+x^{2}}}}}-20\\&={\frac {40x-20{\sqrt {36+x^{2}}}}{\sqrt {36+x^{2}}}}\end{aligned}}$

Setting the derivative to 0 yields

${\begin{aligned}0&={\frac {40x-20{\sqrt {36+x^{2}}}}{\sqrt {36+x^{2}}}}\\0&=40x-20{\sqrt {36+x^{2}}}\\{\sqrt {36+x^{2}}}&=2x\\36+x^{2}&=4x^{2}\\36&=3x^{2}\\12&=x^{2}\\\pm 2{\sqrt {3}}&=x\end{aligned}}$

Notice that the negative answer is inadmissable since it corresponds to a side length (and x should be positive to minimize cost). Now, the endpoints are x = 0 and x = 8. Then,

$\displaystyle C(0)=40{\sqrt {36+0^{2}}}+20(8-0)=240+160=400$

${\begin{aligned}C(2{\sqrt {3}})&=40{\sqrt {36+(2{\sqrt {3}})^{2}}}+20(8-2{\sqrt {3}})\\&=40{\sqrt {36+12}}+160-40{\sqrt {3}}\\&=160{\sqrt {3}}+160-40{\sqrt {3}}\\&=120{\sqrt {3}}+160<400\end{aligned}}$

$\displaystyle C(8)=40{\sqrt {36+8^{2}}}+20(8-8)=40{\sqrt {100}}=400$

and so the minimum occurs at $\displaystyle x=2{\sqrt {3}}$