Question

Find the greatest possible value of the product xy, given that x and y are both positive and x + 2y = 30

Answer 1

greatest value of xy
x + 2y = 30, x = 30 - 2y
xy = (30-2y)y = 30y-2y^2
so if -2y^2+30y = 100, then we can take out a (-2), so: y^2-15y=-50, y^2-15y+50=0
(y-5)(y-10)=0
y-5=0, y=5
y-10=0, y=10
Now we can "plug n play": plug each into x = 30-2y
x = 30-2(5) = 30-10 = 20
x = 30-2(10) = 30-20 = 10
So if we make x=20, y=5... xy = 100
And if x=10, y=10... xy = 100
Now we need to try other number variations to see if they can make xy > 100
if y=6, then x = 30-2 (6) = 30-12 = 18
xy = (18)(6) = 108. Ooh! there's a better choice, but let's keep going
if y=7, then x = 30-2(7) = 16, xy = 7*16 = 112
Even higher! But let's check others
If y=8, then x = 30-16=14, so xy = 112
But from here, the higher y gets the product of xy starts to decrease again
So the greatest possible value comes from 7*16 and 8*14, in which x*y was the maximum it could be at 112

Answer 2

Answer: The maximum value of the product x*y is 112.5

Explanation:

We have that x + 2y = 30

so we can isolate one of the variables and get:

x = 30 - 2y

now, the product of x*y can be written as:

x*y = (30 - 2y)*y = 30y - 2y^2

now, we want to find the maximum of 30y - 2y^2

because the leading coeficient is negative (-2), we know that the hands of the quadratic function will go downwards, so the zero of the derivate of this quadratic equation will give us the value where the equation has a maximum:

f(y) = -2*y^2 + 30*y

f'(y) = -4*y + 30 = 0

y = -30/4 = 7.5

So the maximum value is:

f(7.5) = -2*7.5^2 + 30*7.5 = 112.5