Find two positive numbers whose product is a maximum and whose sum of the first number and four times the second is 120

Answer 1

Hi Bella

You are trying to find two positive x and y numbers such that

x + 4y = 120

And

xy = maximum

You can use a substitution which will give a quadratic to work with since

x + 4y = 120

Then

4y = 120 -x

Divide both sides of the equation by 4 to solve for y

y = 30 - x/4

So substituting

xy = x(30 - x/4)

xy = 30x - x²/4

You have a quadratic in standard form where c=0, c has no value so it is not shown and your maximum is for a parabola

I'll leave it to you to research that a few websites call it the tricky one but looks like this

-x²/4 + 30x + 0 =0

With it set equal to zero you can find x intercepts, you can find the x coordinate of the vertex and we can use it to find the y value and to find your two positive numbers

Now there are a couple of ways to approach this, in the standard form

-x²/4 + 30x =0

a = -1/4

b = 30

c = 0

-b/2a = the x coordinate of the vertex

-b/2a = -30/(2(-1/4))= 30/(1/2) = 60 you can use decimal if its easier for you

x = 60

You can substitute this number in

x + 4y = 120 to quickly solve for y

60 + 4(y) = 120

Subtract 60 from both sides

4y = 60

Divide both sides by 4

y = 15

So your two positive numbers are

x = 60 and y = 15

Your maximum for your quadratic we had above is xy

xy =60(15)= 900

To find the y coordinate you just plug in 60 for x in your quadratic equation

-x²/4 + 30x =y

-60²/4 + 30(60) = y

-3600/4 + 1800 = y

-900 + 1800 = y

900 = y coordinate of the vertex

so (60, 900) is your maximum

Now let's check everything

x + 4y = 120

60 + 4(15) = 120

60 + 60 = 120

And

xy = maximum for the quadratic -x²/4 + 30x =y

60(15) = 900

You can graph

-x²/4 + 30x =y

at desmos.com

You can factor the equation

-x²/4 + 30x =0

x(-x/4 + 30)=0

x =0

-x/4 + 30 = 0

-x/4 = -30

Multiply both sides by -4

x = 120

x intercepts are (0,0) and (120,0)

You can use derivative tests; take the first derivative and set that equal to zero

-x²/4 + 30x

d/dx ( use the constant muliple rule and the power rule)

d/dx(cf(x)) = c(f'x)

-x/2 + 30 = 0

-x/2 = -30

Multiply both sides by -2

x = 60

Check these out you can also use the quadratic equation if you want. Do grah it to check the vertex, and x intercepts.

Answer 2

Bella

I will model this for you, assuming the sum is 200, then you can use my model.

Suppose first number is x and second number y. x and y are both positive.

x + 4y = 200

We want to maximise P = xy

P = xy

x = 200 - 4y

P = y (200 - 4y) = 200y - 4y²

To find the value of y giving a maximum / minimum, we solve dP/dy = 0

Note that P is a quadratic in y, with a < 0 (facing downwards) so the turning point will be a maximum

dP/dy = 200 - 8y

200 - 8y = 0

y = 200 / 8 = 25,

So y will be 25 and x = 200 - 4y = 100, so the maximum product will be 2500

Now follow my model using 120 instead of 200.

Mike