Monday Math 151

Find the values of x,y,z≥0 that maximize the value of f(x,y,z)=xyz, subject to the constraint ax+by+cz=d, where a, b, c and d are positive real numbers.

This is a maximization subject to constraint problem, and is thus best approached by the method of Lagrange multipliers. Thus, we find the stationary points of the function:
(where we have chosen –λ rather than +λ to make some later algebra easier).
Taking the partial derivatives of F, we see
Setting the first three of these equal to zero, we obtain
And setting the fourth equal to zero gives us our constraint ax+by+cz=d, as expected.
Now, multiplying those three equations we obtained from setting the partial derivatives equal to zero, we obtain
and taking the square root of both sides (with λ≥0),
Dividing this by the first equation, we see
And analogously,
And thus, ax=by=cz=\sqrt{\lambda{abc}}
and from our constraint ax+by+cz=d, we see that
(from which we can solve for λ, but this is unnecessary).
Thus, the point which maximizes f(x,y,z)=xyz is (x,y,z)=\left(\frac{d}{3a},\frac{d}{3b},\frac{d}{3c}\right), with the maximum value thus being \frac{d^3}{27abc}.


Tags: , ,

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: