Answer to Question #136821 in Algebra for Cameron Foster

Question

A contracter builds boathouses in two basic models, the Atlantic and the Pacific. Each Atlantic model
requires 1000 ft of framing lumber, 3000 ft3 of concrete and $2000 for advertising. Each Pacific model
requires 2000 ft of framing lumber, 3000 ft3 of concrete and $3000 for advertising. Contracts call for
using at least 8000 ft of framing lumber, 18000 ft3 of concrete, and $15,000 worth of advertising. If the
construction cost fo each Atlantic model is $30,000 and the construction cost for each Pacific model is
$40,000, how many of each model should be built to minimize construction costs? What is the lowest
cost

Answer 1

$$Solution$$

Step 1

Given

A contractor builds boathouses in two basic models, the and the Pacific. Each Atlantic model requires 1000 ft of framing lumber, 3000$ft^3$   of concrete, and $2000 for advertising. Each Pacific model requires 2000 ft of framing lumber, 3000 of concrete, and $3000$ft^3$ for advertising. Contracts call for using at least 8000 ft of framing lumber, 18,000 $ft^3$  of concrete, and $15,000 worth of advertising. If the construction cost for each Atlantic model is $30,000 and the construction cost for each Pacific model is $40,000.

Let x = the number of Atlantic boathouses.

y = the number of Pacific boathouses.

Step 2

Constraints:

(i). Framing lumber:

The Atlantic model requires 1000 ft of framing lumber and the Pacific model requires 2000 ft of framing lumber. We need to use at least 8000 ft of framing lumber. Hence the constraint would be

$$1000x+2000y \geq 8000$$

(ii). Concrete:

The Atlantic model requires 3000 ft³ of concrete and the Pacific model requires 3000 ft³ of concrete. We need to use at least 18,000 ft³ of concrete. Hence the constraint would be

$$3000x+3000y \geq 18000$$

(iii). Advertising: The Atlantic model requires $2000 for advertising and the Pacific model requires $3000 for advertising. We need to use at least $15,000 for advertising. Hence the constraint would be

$$2000x+3000y \geq 15000$$

$x \geq 0\\ y\geq 0$

Step 3

Forming the objective function:

The cost of construction for each Atlantic model is $30,000 while it is $40,000 for each Pacific model. We need to minimize the construction cost. Hence the objective function is

Minimize: $z=30000x + 40000y$

Step 4

We will now change linear inequalities to linear equations by changing ‘’ to ‘=’.

(To plot the graph)

Because of ‘=’ portion of ‘$\leq$ ’ the points of the linear equation satisfy the linear

Inequality and are the part of graph.

$1000x+2000y=8000\\ 3000x+3000y=18000\\ 2000x+3000y=15000$

Step 5

Graph the linear equations as solid lines and obtain a feasible region quadrant I.

Step 6

The corner points $(0,6)$  and $(8,0)$ can be identified from the graphs.

The coordinates of the corner point $(3,3)$  can be found by solving the system

$3000x+3000y=18000\\ 2000x+3000y=15000$

The coordinates of the point (6, 1) can be found by solving the system

$1000x+2000y=8000\\ 2000x+3000y=15000$

Step 7

Use the corner points obtained to solve the objective function.

Atlantic boathouses and Pacific boathouses are building.

Minimum Cost is 210000