# What is the largest area for $800 #### Hypnos_16 1. Homework Statement a rectangular lot adjacent to a highway is to be enclosed by a fence. If fencing cost$6 per metre along the highway and $4 per metre on the other three sides, find the largest area that can be fenced off for$800.

I get these questions generally, because they're usually area problems, but when they tossed in cost, and that it's the same on 3 walls yet different on the fourth it's just throwing me off.

2. Homework Equations

I get that you have two equations in this case something like

xy = Area
x + 3y = 800

(I don't think they're the right ones, just it's supposed to be something like that, if they are right though let me know)

3. The Attempt at a Solution

I haven't tried it cause i don't know how to start it

Related Calculus and Beyond Homework Help News on Phys.org

#### lanedance

Homework Helper
you equations aren't right
so say
x = width (parallel to highway)
y = length (perp to highway)
perimeter length = 2x + 2y

now try and write down the cost, you have x at $6/m and 2y+x at$4/m

Last edited:

#### Hypnos_16

so if x = $6/m and 2y + x =$3 / m
then
6(x) + 3(2y + x) = 800
6x + 6y + 3x = 800
9x + 6y = 800
x = (800 - 6y) / 9
something like that?

#### armolinasf

I'm not seeing the logic behind 2y + x = $3 / m If only one side is facing the highway at a cost of 6$ and the other three are 4$, you would have something like (2x*4$)+6$y+4$y=800\$.

#### lanedance

Homework Helper
sorry meant 4

so you get
800 = 6x + 4(2y+x) = 10x + 8y

now rearange for y in terms of x, y(x) and substitue into your area A(x,y)
A(x) = A(x, y(x))

then maximise in terms of x

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving