A landscape architect wishes to enclose a rectangular garden on one side by a brick wall costing $12 per foot and on the other three sides by a metal fence costing $4 per foot. If the area of the garden is 200 ft, find the dimensions of the garden minimizing the cost. (Let x be the length of the brick wall and y be the length of an adjacent side in feet.)

The dimensions of the garden that will minimize the costs are: x=10ft, y=20ft

Step-by-step explanation:

Let x be the length of the brick wall. Let y be the length of the adjacent sides.

Area of the garden = 200 ft

Therefore: xy=200

Perimeter of the Fence = Length of

brick wall+Length of Opposite Metal Fence + Length of Other Adjacent Metal Fence

=x + x + 2y

Costing of brick wall= $12 per foot Cost of Metal =$4 per foot.

Cost of the Fence=12x+4x+4(2y)

=16x+8y

Recall:

Therefore, the cost in terms of x is:

The minimum costs C(x) occur when its derivative equals zero.

Thus:

x=10ft

Recall that:

y=200/10=20 ft

The dimensions of the garden that will minimize the costs are: x=10ft, y=20ft

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

exa:

)

3+11

14

step-by-step explanation:

from this example, we can see that subtracting with integers is the same as adding the opposite. we can solve using a number line or by using the saying "keep, change, change" to us solve. it is important to remember that we can subtract a negative and end up with a larger number than we started with.