# Thread: differentiation: maxima and minima

1. ## differentiation: maxima and minima

p80 q11 ex18c
question:
A and D are two towns on the same side of the main road BC. A road APD is to be made to join A and D to the main road. where should the location of P be in order that the length of the road is its smallest? you can only use differentiation method.

my working:
let $\displaystyle l$ be the length of the road, x be the distance of BP
$\displaystyle l = \sqrt {x^2+4} +\sqrt {(10-x)^2+9}$
$\displaystyle dl/dx = \frac 1 2 (x^2+4)^{-\frac 1 2 } (2x) + \frac 1 2 (x^2 -20 x +109)^{-\frac 1 2 } (2x-20)$
$\displaystyle = x(x^2+4)^{-\frac 1 2 }+ (x-10)(x^2 - 20 x +109)^{-\frac 1 2 }$
i cannot find the value of x when dl/dx = 0 . thanks!

2. Hello,

Originally Posted by afeasfaerw23231233
p80 q11 ex18c
question:
A and D are two towns on the same side of the main road BC. A road APD is to be made to join A and D to the main road. where should the location of P be in order that the length of the road is its smallest? you can only use differentiation method.

my working:
let $\displaystyle l$ be the length of the road, x be the distance of BP
$\displaystyle l = \sqrt {x^2+4} +\sqrt {(10-x)^2+9}$
$\displaystyle dl/dx = \frac 1 2 (x^2+4)^{-\frac 1 2 } (2x) + \frac 1 2 (x^2 -20 x +109)^{-\frac 1 2 } (2x-20)$
$\displaystyle = x(x^2+4)^{-\frac 1 2 }+ (x-10)(x^2 - 20 x +109)^{-\frac 1 2 }$
i cannot find the value of x when dl/dx = 0 . thanks!
$\displaystyle \frac{dl}{dx}=\frac{x}{\sqrt{x^2+4}}+\frac{x-10}{\sqrt{x^2-20x+109}}$

with x between 0 and 10.

Put on the same denominator :

$\displaystyle \frac{dl}{dx}=\frac{x \sqrt{x^2-20x+109}+(x-10) \sqrt{x^2+4}}{\sqrt{x^2-20x+109} \sqrt{x^2+4}}$

The denominator never annulates, so it's ok;

$\displaystyle x \sqrt{x^2-20x+109}+(x-10) \sqrt{x^2+4}=0$

$\displaystyle x \sqrt{x^2-20x+109}=(10-x) \sqrt{x^2+4}$

Squaring the equation :

$\displaystyle x^2 (x^2-20x+109)=(100-20x+x^2)(x^2+4)$

$\displaystyle {\color{blue}x^4}{\color{red}-20x^3}+109x^2=100x^2+400{\color{red}-20x^3}-80x+{\color{blue}x^4}+4x^2$

$\displaystyle 109x^2=100x^2+400-80x+4x^2$

$\displaystyle x^2+16x-80=0$

Can you solve it from here ? (don't forget that x is between 0 and 10)