-the sum of one number and four times another number is 64 Find the value of the numbers so that their product is a maximum?

Answer 1

The numbers are 8 and 32 and the max product is 256.

Let one number be m
Let the other be n
We have m + 4n = 64
So 4n = 64 - m
or n = 16- m/4

We want mn = m(16 - m/4) to be a max value.

That is to say the product of these two numbers equals -(1/4)m2 + 16m.

Now depending on you level of math there are many ways to do this.

If you know calculus, you can take the derivative of f(m)= -(1/4)m2 + 16m
and you find it as -(1/2)m + 16.
Now you would set that equal to zero which will indicates m = 32.

So you have:

-(1/2)m + 16 = 0
m = 32

To find the other number, substitute 32 for m into the equation n = 16 - m/4 and solve for n.
So that the other number is 16-32/4 or 8.
Thus, the numbers are 8 and 32, and their product is 256.
Since f(m)=16m-m2 /4, we can also look at f(32)= 16(32)-322 /4=512-256=256

METHOD TWO

Now if you don't know calculus, here is another way to do it.
You can see that f(m)= -(1/4)m2 + 16m is a quadratic function written in standard form as
ax2 + bx + c, where a = -1/4, b = 16, and c = 0.
The graph is a parabola which opens down since the sign of the coefficient of m2 is negative (a = -1). We need to find the vertex of the parabola where the y-coordinate will be the max value.

The formula for the vertex is (-b/2a, f(-b/2a)), so we have

-b/2a = -16/(2(-1/4)) = 32, and
f(m) = -(1/4)m2 + 16m
f(-b/2a) = f(32) = -(1/4)(32)2 + 16(32) = 256

Therefore, the vertex of the parabola is at (32, 256) and the maximum value of 256 happens when m = 32. Since this max value is the product of m and n, then n = 8 (256/32).



METHOD THREE

Once again look at the function f(m)=16m-m2 /4 and write it in standard form
f(m)=-m2 /4 +16m
Now complete write this as -1/4(m2 -64m) and complete the square.
We have
f(m)=-1/4(m -32)2 +256
This tells us the graph is a parabola with vertex (32, 256)
Since the parabola opens downward, 256 is the max.