There are 400 available to fence in a rectangular garden The

There are $400 available to fence in a rectangular garden. The fencing for the side of the garden facing the road costs $15 per foot, and the fencing for the other three sides costs $5 per foot. Consider the problem of finding the dimensions of the largest possible garden.

(a) what would be the objective equation and what would be the constraint equations?

(b) Express the quantity to be maximized as a function of x.

(c) Find the optimal values of x and y.

Solution

a)The objective is to maximize area and so,
objective equation: A = xy
The constraint is the amount of money we have for the fence, i.e. we only have 400 dollars
constraint equation: 400 =(cost of fence parallel to highway) +(cost of 3 other sides)
400 = 15x + 5(5y + x)
400 = 20x + 25y

b)We use the constraint equation to solve for y:

400 = 20x + 25y =) y = (400-20x)/25=16-(4/5)x
Now, plug this y into our objective equation to nd the quantity we want to maximize as
a function of x:
A(x) = x(16-(4/5)x) = 16x -4(x^2)/5

c)Find the critical value of A(x):
dA/dx = 16-8x/5 = 0

=) 16=8x/5

8x=80

x=10
Now, we need to make sure that this value is the maximum using the second derivative:

d^2A/dx^2 = -8/5

=) d^A/dx^2(10) < 0

not only is there a relative maximum at x = 10, but we can conclude that the
maximum occurs as x = 10 since d^A/dx^2< 0 for all x. We also need y:
y = 16-(4/5)x

=) y = 16-(4/5)*10

= 16-8

y=8