1. ## Area Word Problems

1) A rectangular garden is bounded on one side by a river and is to be fenced in on the three other sides. Fencing material for the side parallel to the river is $30 per foot. Materials for the two other sides cost$10 per foot. What are the dimensions of the garden of largest possible area if $1200 is to be spent for fencing material? 2) A rectangular piece of cardboard measuring 20 inches by 32 inches is to be made into a box with an open top by cutting equal size squares from each corner and folding up the sides. Let x represent the length of a side of each square. What is the maximum volume of this box? 2. 1) may be not enough info. 2) (20 - 2x)(32 - 2x)(x) cubic inches 3. Originally Posted by NRS11 1) A rectangular garden is bounded on one side by a river and is to be fenced in on the three other sides. Fencing material for the side parallel to the river is$30 per foot. Materials for the two other sides cost $10 per foot. What are the dimensions of the garden of largest possible area if$1200 is to be spent for fencing material?

2) A rectangular piece of cardboard measuring 20 inches by 32 inches is to be made into a box with an open top by cutting equal size squares from each corner and folding up the sides. Let x represent the length of a side of each square. What is the maximum volume of this box?
1) let the distance of the fence opposite to the river be x and that of the adjacent side be y. Then 30x+20y=1200, isolate x, and the area is x*y. You only have to complete the square after i think

4. Originally Posted by NRS11
1) A rectangular garden is bounded on one side by a river and is to be fenced in on the three other sides. Fencing material for the side parallel to the river is $30 per foot. Materials for the two other sides cost$10 per foot. What are the dimensions of the garden of largest possible area if $1200 is to be spent for fencing material? ... Let r be the distance of the fence parallel to the river. Let s be the length of the fence perpendicular to the river. Max Money:$1200 = $30*r + 2*$10*s
Max Area: r*s

$1200 = 30r + 20s$

isolate r
$\dfrac{1200 - 20s}{30} = r$

substituting that in the area equation

$\text{Area = } \left ( \dfrac{ 1200 - 20s}{30} \right ) s$

$\text{Area = } 40s - \dfrac{ 2s^2}{3}$

take the derivitive and determine the value of s when the slope is zero.

$0 = 40 - \dfrac{ 4s }{3}$

solve that to get the value of s.
plug that back into the equation above to compute r.

How that helps

.