Question

Two industrial plants, 𝐴𝐴 and 𝐵𝐵, are located 15 miles per apart and emit 75 ppm (parts per

million) and 300 ppm of particular matter, respectively. Each part is surrounded by a restricted

area of radius 1 mile in which no housing is allowed, and the concentration of pollutant arriving

at any other point 𝑄𝑄 from each plant decreases with the reciprocal of the distance between that

plant and 𝑄𝑄. Where should a house be located on a road joining the two plants to minimize the

total pollution arriving from both plants?

Answer 1

Let $x$ denote the distance between the house and Plant A in miles. Then the distance between the house and Plant B will be $15-x$ miles.

Also, since Plant A has a restricted area of 1 mile around it and Plant B has a restricted area of 1 mile around it, therefore $1\leq x \leq14.$

The smog from plant A is $\dfrac{75}{x}.$ The smog from plant B is $\dfrac{300}{15-x}.$

Total pollution arriving from both plants is

$$f(x)=\dfrac{75}{x}+\dfrac{300}{15-x}, 1\leq x \leq14$$

Find the first derivative with respeect to $x$

$$f'(x)=(\dfrac{75}{x}+\dfrac{300}{15-x})'$$

$$=-\dfrac{75}{x^2}+\dfrac{300}{(15-x)^2}$$

Find the critical number(s)

$$f'(x)=0=>-\dfrac{75}{x^2}+\dfrac{300}{(15-x)^2}=0$$

$$4x^2=(15-x)^2$$

$$3x^2+30x-225=0$$

$$x^2+10x-75=0$$

$$(x+5)^2=100$$

$$x_1=-15, x_2=5$$

Critical numbers: $-15, 5.$

Since $1\leq x\leq 14,$ we take $5$ as the critical number.

$$f(1)=\dfrac{75}{1}+\dfrac{300}{15-1}=96\dfrac{3}{7}$$

$$f(14)=\dfrac{75}{14}+\dfrac{300}{15-14}=305\dfrac{5}{14}$$

$$f(5)=\dfrac{75}{5}+\dfrac{300}{15-5}=45$$

$$f(1)<f(5)<f(14).$$

Therefore, the house should be located 5 miles from plant A.