The graph shows equivalent ratios. Frank plotted another point with a y-value of 18. What is the x-value of that point?
1 answer:
You might be interested in
Answer:
x = 0.63 miles
y = 3.10 miles
Step-by-step explanation: See Annex
Vehicle capacities.
In land 1 mile in 5 minutes then 1/5 mile in 1 minute
In water 1 mile in 4 minutes then 1/4 mile in 1 minute
Let call x the number of miles in water and y the number of miles in land
Then the route will be x (miles in water ) and y miles in land
From Annex
L = 1/4*x + 1/5*y
From Figure (Annex) y = √( 3 - x )² + ( 2)² Pithagoream theorem
y = √ 9 + x² -6*x + 4 ⇒ y = √ x² - 6*x + 13
Then
L(x) = 1/4x + 1/5√ x² - 6*x + 13
Taking derivatives on both sides of the equation:
L´(x) = 1/4 + 1/5* 1/2 (2*x -6) / √ x² - 6*x + 13
L´(x) = 0 ⇒ 1/4 + 1/5* 1/2 (2*x -6) / √ x² - 6*x + 13 = 0
[1/4* √( x² - 6*x + 13) + 1/5* (x -3) ] /√ x² - 6*x + 13 = 0
1/4* √( x² - 6*x + 13) + 1/5*x - 3/15 = 0 ⇒ 1/4* √( x² - 6*x + 13) = 1/5 ( x -3)
1/4* √( x² - 6*x + 13) = 1/5*x - 1/5
Squaring both sides
1/16* (x² - 6*x + 13) = 1/25 (x - 3 )² ⇒ x²/16 - 6/16 *x + 13/16 = 1/25 ( x² + 9 - 6*x
25 (x² - 6*x + 13) = 16 ( x² + 9 -6*x)
25*x² - 150*x + 325 = 16x² + 144 - 96*x
9*x² - 294*x + 181 = 0
x₁,₂ = 294 ± √ (86436 - 6516) / 18
x₁ = 294 + 282,7 / 18 we dismiss that solution si not feasible with problem statemente
x₂ = 294 - 282,7 / 18
x₂ = 11.3/ 18
x₂ = 0.63 miles and y = √ x² - 6*x + 13 ⇒ y = 3.10 miles
