An electronics firm makes a clock radio in two different...

An electronics firm makes a clock radio in two different models: one (model 380 ) with a battery backup feature and the other (model 360 ) without. It takes 1 hour and 15 minutes to manufacture each unit of the model 380 radio, and only 1 hour to manufacture each unit of the model $360 .$ At least 500 units of the model 360 radio are to be produced. The manufacturer realizes a profit per radio of $\$ 15$ for the model 380 and only $\$ 10$ for the model $360 .$ If at most 2000 hours are to be allocated to the manufacture of the two models combined, how many of each model should be made to maximize the total profit?

Key Concepts

Corner Point Method

The corner point method is a technique used in linear programming where the objective function is evaluated at the vertices (corner points) of the feasible region. The optimal solution is found at one of these points, making this a key method for solving such problems.

Feasible Region

The feasible region is the set of all points that satisfy all of the constraints in a linear programming problem. It represents all possible solutions, and the optimal solution is typically found on the boundary, often at a vertex of this region.

Constraints

Constraints are the inequalities or equations that represent the limitations or requirements in a linear programming problem. They restrict the values that the decision variables can take and form the boundaries of the feasible region.

Objective Function

The objective function is a linear equation that represents the goal of the optimization, such as maximizing profit or minimizing cost. It is constructed in terms of decision variables and is the expression that is to be maximized or minimized in the problem.

Linear Programming

Linear programming is an optimization method used to determine the best outcome, such as maximum profit or minimum cost, under a set of linear constraints. It involves formulating a problem with an objective function and a set of linear inequalities or equalities representing limitations or requirements.

Transcript

00:01 Here in this problem we have to maximize the total profit of an electronic firm and it is given that an electronic firm manufacture two type of model first one is model 380 and the second one is model 360 and it takes one hour 15 minutes to manufacture one unit of model 380 so one hour 15 minute means 75 minutes and it takes one hour to manufacture model 360 that means 60 minutes and we have to manufacture at least 500 unit of the model 360 so let us assume that we are manufacturing x unit of model 380 and y unit of model 360 and it is given it the manufacturer obtains a profit of $15 for the one unit of model 380 and $10 for the one unit of model 360 so we have to maximize the profit p is equals to 15 x plus 10 y and the constraint are that y should be greater than or equals to 500 and the next is we are available with the 2 ,000 hours to manufacture both the product so 2 ,000 hours mean 120 ,000 minutes so we can write the next inequality as 75x plus 60 y should be less than or equals to 120 ,000 so these are the two inequality that we are able to obtain for the given statement now i'll be graphing these inequality on a graph paper so let's take a graph paper now so this is the y axis and this is the x -axis now this line represents the equation 75x plus 60 y is equals to 120 ,000 and this line represents the equation y equals to 500 and the common region for the obtained inequality would be this region and there are three corner points now we have to check the value of p at each and every corner points for that we have to obtain the coordinates so coordinate of this point is 0 2000 and the coordinate of this point is 0 500 similarly here we get to coordinate as twelve hundred comma five hundred now we have to obtain the value of p at each and every coordinate so p here comes out to be as twenty thousand i've just plugged the value of x as zero and y as two thousand so i get the value of p as twenty thousand similarly here we get the value of p as five thousand and here we get the value of p as twenty three thousand so this is the maximum value of p that we obtained at this coordinate so x comes out to be as 1200 and y comes out to be as 500...

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