MHB Maximizing Production Output with a Cost of 10

[evinda]

Hello! (Wave)

The production function of a company is $Q(x,y)=xy$. The cost of the production is $C(x,y)=2x+3y$.
If this company can spend $C(x,y)=10$, which is the greatest quantity that it can produce?We want to maximize $Q(x,y)=xy$ under the condition $2x+3y=10$.

We use the method of Lagrange multipliers. The restriction is $g(x,y)=2x+3y-10=0$.

So we are looking for $\lambda, x ,y$ such that $\nabla Q(x,y)= \lambda \nabla g(x,y)$ and $g(x,y)=0$.

So $y= 2 \lambda, x=3 \lambda \Rightarrow 2x=3y$.

So we get that $x=\frac{10}{4}, y=\frac{5}{3}$.

So the point $\left( \frac{10}{4}, \frac{5}{3}\right)$ is an extremum. How do we deduce that it is maximum?

 

[I like Serena]

Hello! (Wave)

The production function of a company is $Q(x,y)=xy$. The cost of the production is $C(x,y)=2x+3y$.
If this company can spend $C(x,y)=10$, which is the greatest quantity that it can produce?We want to maximize $Q(x,y)=xy$ under the condition $2x+3y=10$.

We use the method of Lagrange multipliers. The restriction is $g(x,y)=2x+3y-10=0$.

So we are looking for $\lambda, x ,y$ such that $\nabla Q(x,y)= \lambda \nabla g(x,y)$ and $g(x,y)=0$.

So $y= 2 \lambda, x=3 \lambda \Rightarrow 2x=3y$.

So we get that $x=\frac{10}{4}, y=\frac{5}{3}$.

So the point $\left( \frac{10}{4}, \frac{5}{3}\right)$ is an extremum. How do we deduce that it is maximum?

Hey evinda! (Smile)

The standard method is the second derivative test using the Hessian matrix. (Thinking)

However, we can make it a bit easier for ourselves.
Since we only have 1 extremum of a function that is continuously differentiable, we can inspect any other point and see if it is more or less than our extremum.
Suppose we pick $(x,y) = (5,0)$, which satisfies $g(x,y)=0$, then we get the quantity $Q(5,0)=5\cdot 0 = 0$.
This is less than the extremum we found (which is at least positive), therefore it is a maximum. (Emo)

 

[evinda]

Hey evinda! (Smile)

The standard method is the second derivative test using the Hessian matrix. (Thinking)

But in our case it does not hold that $Q_x{\left( \frac{10}{4}, \frac{5}{3}\right)}=Q_y{\left( \frac{10}{4}, \frac{5}{3}\right)}=0$. Does it? (Sweating)

However, we can make it a bit easier for ourselves.
Since we only have 1 extremum of a function that is continuously differentiable, we can inspect any other point and see if it is more or less than our extremum.
Suppose we pick $(x,y) = (5,0)$, which satisfies $g(x,y)=0$, then we get the quantity $Q(5,0)=5\cdot 0 = 0$.
This is less than the extremum we found (which is at least positive), therefore it is a maximum. (Emo)

So we pick any point satisfying $g(x,y)=0$, right? How do we use the fact that the function is continuously differentiable?

 

[I like Serena]

But in our case it does not hold that $Q_x{\left( \frac{10}{4}, \frac{5}{3}\right)}=Q_y{\left( \frac{10}{4}, \frac{5}{3}\right)}=0$. Does it? (Sweating)

Indeed, that is because we have the boundary condition $g(x,y)=0$ that we're not taking into account now.
We are actually maximizing $\Lambda(x,y,\lambda) = Q(x,y) - \lambda g(x,y)$.
It means that we need to apply the second derivative test to $\Lambda$.
See example 2 in Lagrange Multiplier.

So we pick any point satisfying $g(x,y)=0$, right? How do we use the fact that the function is continuously differentiable?

Hold on. (Wait)
It is actually a little more complicated since the extremum could be a saddle point.
In that case it does not suffice to check a single distinct point.
So we need the second derivative test after all. (Tmi)

 

[evinda]

Indeed, that is because we have the boundary condition $g(x,y)=0$ that we're not taking into account now.
We are actually maximizing $\Lambda(x,y,\lambda) = Q(x,y) - \lambda g(x,y)$.
It means that we need to apply the second derivative test to $\Lambda$.
See example 2 in Lagrange Multiplier.

But $Q_{xx}(x,y)=0$ so we cannot apply the theorem. Or am I wrong?

Since $D$ is negative, we deduce that it is a saddle point. But this is not what we want to show... (Sweating)

 

[I like Serena]

But $Q_{xx}(x,y)=0$ so we cannot apply the theorem. Or am I wrong?

Since $D$ is negative, we deduce that it is a saddle point. But this is not what we want to show... (Sweating)

We need to apply it to $\Lambda(x,y, \frac 56)$... (Thinking)

 

[evinda]

We need to apply it to $\Lambda(x,y, \frac 56)$... (Thinking)

Isn't this equal to $xy-\frac{10}{6} x-\frac{5}{2} y+\frac{50}{6}$?

But then again $\Lambda_{xx}=0$. Or am I wrong?

 

[I like Serena]

Isn't this equal to $xy-\frac{10}{6} x-\frac{5}{2} y+\frac{50}{6}$?

But then again $\Lambda_{xx}=0$. Or am I wrong?

That's right. It means that the test is inconclusive. Alternatively, we can make a drawing from which it will be evident. (Thinking)
View attachment 5614

We can use the Extreme Value Theorem to prove it.
Suppose we bound the domain to, say, $[0,5]\times[0,5]$.
Then we can find all extremes either at the boundary, or where the gradient is zero (since the function is continuously differentiable).
The boundary consists of only 2 points at $(5,0)$ respectively $(0,3\frac 13)$, due to the restriction $g(x,y)=0$.
In both cases we have $Q(x,y)=0$.
And we found the extremum $Q(\frac 52,\frac 53) = \frac{25}{6} > 0$.
So on our bounded domain, the Extreme Value Theorem tells us that we must have a maximum at $(\frac 52,\frac 53)$, since all other extremes are lower.

If we increase the size of our domain, we will always have exactly 2 boundary points.
Checking them, we'll see that $Q < 0$, meaning that our conclusion does not change. (Emo)
Yet another alternative is to use a different approach by substituting the boundary condition into the Q(x,y).
That is:
$$y=\frac 13(10-2x) \\ Q = x \cdot \frac 13(10-2x)$$
Now we can find the extrema of Q in the regular way and use the second derivative to find that it's a maximum. (Thinking)