# Thread: Problem in obtaining a solution

1. ## Problem in obtaining a solution

I have a problem in finding the solution of problem. The problem is as follows:

Find the maximum value of yz + xz subject to the two constraints, y + 2z = 1, x + z = 3.

The answer in the textbook is 4/3. However, the answer I have is -(3/4)

Is my answer wrong? if wrong, how can I get to the answer 4/3?

Thank you very much

2. Originally Posted by Real9999
I have a problem in finding the solution of problem. The problem is as follows:

Find the maximum value of yz + xz subject to the two constraints, y + 2z = 1, x + z = 3.

The answer in the textbook is 4/3. However, the answer I have is -(3/4)

Is my answer wrong? if wrong, how can I get to the answer 4/3?

Thank you very much
You need to express x and y in terms of z first.
Then evaluate yz+xz in terms of z.
You get a quadratic in z, with the co-efficient of the square being negative.
Hence the quadratic in z is an inverted U-shape.
The derivative of the quadratic being zero (horizontal tangent resting on top of the curve) locates z giving max yz+xz.

$\displaystyle y+2z=1\ \Rightarrow\ y=1-2z$

$\displaystyle x+z=3\ \Rightarrow\ x=3-z$

Then

$\displaystyle yz+xz=(1-2z)z+(3-z)z$

Evaluate this and differentiate (then set derivative to zero) to find z that gives the maximum yz+xz.

Substitute in this value of z to find the maximum value of yz+xz.

3. Yes, your answer is wrong and the answer in the book is correct. Unfortunately, since you did not say how you got -3/4, I don't see how anyone can tell where you made a mistake.

Not showing how you got a wrong answer also means we do not know what method would be appropriate for your level of knowledge. Archie Meade's method is the most elementary- you don't even need to differentiate- since the equation is z is quadratic, you can find the value of z that gives the maximum by completing the square.

If you are advanced enough you could also do this using the Lagarange multiplier method. It is mored sophisticated but the calculations are simpler.

Let f(x, y, z)= yz+ xz be the function you want to maximize (the "target"function) and let g(x, y, z)= y + 2z = 1 and h(x, y, z)= x + z = 3 be the constraints. At a maximum or minimum, the gradient of f will be a linear combination of the gradients of g and h: $\displaystyle \nabla f= \lambda_1\nabla g+ \lambda_2\nabla h$.

For these functions, that says $\displaystyle z\vec{i}+ z\vec{j}+ (x+ y)\vec{k}= \lambda_1(\vec{j}+ 2\vec{k})+ \lambda_2(\vec{i}+ \vec{k})$.

The components give $\displaystyle z= \lambda_2$, $\displaystyle z= \lambda_2$, and $\displaystyle x+ y= 2\lambda_1+ \lambda_2$.

From the first two equations, $\displaystyle \lambda_1= \lambda_2= z$ so the third equation is $\displaystyle x+ y= 3z$.

Now, we have three equations to solve for x, y, and z: x+ y= 3z, y+ 2z= 1, and x+ z= 3. From the third, z= 3- x so the first equation is the same as x+ y= 3(3- x) = 9- 3x or 4x+ y= 9. The second equation is the same as y+ 2(3- x)= y+ 6- 2x= 1 or 2x- y= 5. Adding 4x+ y= 9 and 2x- y= 5 eliminates y leaving 6x= 14 or x= 7/3.

From y= 2x- 5, y= 14/3- 15/3= -1/3 and from x+ z= 3, z= 3- x= 3- 7/3= 9/3- 7/3= 2/3. x= 7/3, y= -1/3, and z= 2/3 is the only critical point.

Putting those into f(x, y, z)= yz+ xz we have f(7/3, -1/3, 2/3)= -2/9+ 14/9= 12/9= 4/3.

4. Thank you so much!