Optimization of Rectangle A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing \$50 per foot and on the other three sides by a metal fence costing \$10 per foot. If the area is 24 square feet, what are the dimensions of the garden that minimize the cost? (Let x= side with bricks; y= adjacent side)
I tried finding the length of one side as 24/x and the other as 24/y, but it 
hasn't gotten me far. I also thought about finding the variables through the perimeter, but I realized all they gave was the area.
Thank you!
 A: Let $C(x)$ be the the cost of covering the garden of a side $x$. Since $y=24/x$ (by area), $C(x)=50x+10x+2(10\frac{24}{x})=60x+\frac{480}{x}$. Since $0<x<240$, our restriction will be $x\in(0,240)$. To find for critical number/s,
$$C'(x)=60-\frac{480}{x^2}=0\ (or\ undefined)$$
$$x^2=\frac{480}{60}=8$$
$$\therefore x=2\sqrt{2}\ or\ -2\sqrt{2}$$but $-2\sqrt{2}\notin(0,240)$, so $x=2\sqrt{2}$. To test if it is a relative extremum:
$$C''(x)=\frac{960}{x^3}$$
$$C''(2\sqrt{2})=\frac{960}{(2\sqrt{2})^3}>0\ (rel. min.)$$
Since, there is only one relative extremum in $(0,240)$, therefore it will the absolute extremum, too.
Therefore, the $x$ (side with bricks) is $2\sqrt{2}ft.$, and the other side is $6\sqrt{2}ft.$ Sorry for a not so good solution, but hope it helps. Sorry my previous answer :(
A: First you have to know that the Area will equal x times y.
A=xy
You know the area is 24 square feet, so you can plug that in for A.
24=xy
From here you solve for y and it is y=42/x
Now you need to figure out a formula to find the cost to build this rectangular garden. 
The formula is C(x)= 50x + 10(x+2y)
The formula represents one side which is $50 and three sides worth $10 each. You have one x side left over and the 2y sides left. You add them because it's for perimeter. 
Next you distribute out the equation and will get C(x)=50x + 10x +20y which will equal 60x + 20y. 
You have what y equals, so you plug that into the equation. Your new equation will look like C(x)= 60x + 20(42/x). Now you have only one variable in the equation. You can distribute more and your new equation is
C(x)= 60x + (840/x)
To solve for x you have to take the derivative of that. The derivative is
C'(x)= 60 - (840/x^2). To solve for x you set it equal to zero. When you solve for x you get square root of 14. That is your x value and to solve for y you plug the square root of 14 into y=42/x. 
