# Min-Max problem

#### Yankel

##### Active member
Hello

I could do with a hint or two for this one:

An squared area is to be bounded by using two types of fences. Two parallel flanks of the area will be bounded using a fence that cost \$60 per meter, while the other two will be bounded with a fence that cost \$50 per meter. What is the area with the maximal size which we can bound like this, so that our total expense will be \$6000 ? Thanks ! Last edited by a moderator: #### MarkFL ##### Administrator Staff member The dollar sign character is used as a delimiter of LaTeX code, so if you wish to use this character to represent dollars, precede it with a backslash "\". I'm assuming "squared area" means rectangular. If we let x be the length of each parallel flank which costs \$60 per meter and y be the length of each parallel flank which costs \$50 per meter, then we require:$2x(60)+2y(50)=6000$Dividing through by 20, we obtain:$6x+5y=300$Now, what is the area, in terms of the two variable, that we wish to maximize? #### Yankel ##### Active member the area should be x*y so are you saying to write y as a function of x, and then use it in x*y, which will be a function of x too, which I will maximize ? the final answer I got is 750, is it correct ? Last edited: #### MarkFL ##### Administrator Staff member Yes, that's right:$\displaystyle A(x,y)=xy\$

Now, you have three ways to proceed.

You may take the constraint, solve for one of the variables, then substitute for that variable into the objective function to get a quadratic objective function in one variable.

i) Pre-Calculus method

You may then choose to find the axis of symmetry which will give you the value of that variable which optimizes the objective function. Observing whether this parabola opens upwards or downwards will be enough to determine if the vertex is at a minimum or maximum. Then evaluate the objective function at the axis of symmetry to determine the optimal area.

ii) Single-variable calculus method

You may choose to differentiate the function, equate to zero to find the value of the variable which optimizes the objective function. Then you may use either the first or second derivative test to determine if you have a maximum or minimum. Then evaluate the objective function at this critical value to determine the optimal area.

iii) Multi-variable calculus method

You may also choose to use Lagrange multipliers to optimize the objective function using partial derivatives. This will tell you the relationship between the two variables, which you can then use in the constraint to determine the actual critical values, which you then use in the objective function to determine the optimal area.

I suspect you are to use the second option. 