<h3>Given</h3>
Two positive numbers x and y such that xy = 192
<h3>Find</h3>
The values that minimize x + 3y
<h3>Solution</h3>
y = 192/x . . . . . solve for y
f(x) = x + 3y
f(x) = x + 3(192/x) . . . . . the function we want to minimize
We can find the x that minimizes of f(x) by setting the derivative of f(x) to zero.
... f'(x) = 1 - 576/x² = 0
... 576 = x² . . . . . . . . . . . . multiply by x², add 576
... √576 = x = 24 . . . . . . . take the square root
... y = 192/24 = 8 . . . . . . . find the value of y using the above equation for y
The first number is 24.
The second number is 8.
Answer:
a 30 pound weight change in 1 month
Step-by-step explanation:
5 pounds
--------------- times 6 months results in a 30 pound weight change in 1 month
1 month
Answer:
13 and 27
Step-by-step explanation:
y= x+14
x+y=40
x+ x+14=40
2x+14=40
40-14=26
so, 2x=26
26/2
=13
so, x=13 and
and y equal 40-13
y=27
