Question

A civil engineer found that the durability d

d

 of the road, she is laying depends on two functions y

1

y1

 and y

2

y2

 as follows: d

=

y

2

2

+

4

y

1

−

21

d=y22+4y1−21

. Functions y

1

y1

 and y

2

y2

 depend on the amount of plastic (x

x

) mixed in bitumen, and their variations are linear functions of x

x

. Let y

1

=

8

y1=8

 and y

2

=

3

y2=3

 for x

=

x=0

 and y

1

=

y1=0

 and y

2

=

7

y2=7

 for x

=

7

x=7

. Find the durability of the road (upto 2 decimal points), if the amount of plastic is such that both the functions are equal.

Answer 1

Let $y_1=ax+b, y_2=mx+n.$

$$y_1(0)=b=8,$$

$$y_2(0)=n=3$$

$$y_1(7)=a(7)+8=0=>a=-\dfrac{8}{7}$$

$$y_2(7)=m(7)+3=7=>m=\dfrac{4}{7}$$

$$y_1(x)=-\dfrac{8}{7}x+8$$

$$y_2(x)=\dfrac{4}{7}x+3$$

If $y_1=y_2,$ then

$$-\dfrac{8}{7}x+8=\dfrac{4}{7}x+3$$

$$\dfrac{12}{7}x=5$$

$$x=\dfrac{35}{12}$$

$$y_1=-\dfrac{8}{7}(\dfrac{35}{12})+8=\dfrac{14}{3}=y_2$$

$$d=(\dfrac{14}{3})^2+4(\dfrac{14}{3})−21=\dfrac{175}{9}\approx19.44$$

$$d=19.44$$