Question

A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol. How Many packets of each food should be used to maximize the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?

Answer 1

Let the dietician uses x packets of food P and y packets of Q.

we construct the following table:

$\begin{array}{cccccc}Foods& Numberof& Amountof& Amoutof& Amountof& Amountof\\ & packets& calcuim& iron& cholesterol& vitaminA\\ P& x& 12x& 4x& 6x& 6x\\ Q& y& 3y& 20y& 4y& 3y\\ Total& x+y& 12x+3y& 4x+20y& 6x+4y& 6x+3y\\ Requires& & Atleast240& Atleast460& Almost300& \end{array}$

So, our problem is to maximize Z = 6x + 3y .....(i)

Subject to constraints $12x+3y\ge 240\iff 4x+y\ge 80$ ......(ii)

$4x+20y\ge 460\iff x+5y\ge 115$ ......(iii)

$6x+4y\le 300\iff 3x+2y\le 150$ ......(iv)

$x\ge 0,y\ge 0$ ......(v)

Firstly, draw the graph of the line 4x + y = 80

$\begin{array}{ccc}x& 20& 0\\ y& 0& 80\end{array}$

Putting (0, 0) in the inequality 4x + y $\ge$ 80, we have

$4\times 0+0\ge 80\Rightarrow 0\ge 80$ (which is false)

So, the half plane is away from the origin.

Secondly, draw the graph of the line x + 5y = 115

$\begin{array}{ccc}x& 0& 115\\ y& 23& 0\end{array}$

Putting (0, 0) in the inequality x + 5y $\ge$ 115, we have

$0+5\times 115\Rightarrow 0\ge 115$ (which is false)

So, the half plane is a way from the origin.

Thirdly, draw the graph of the line 3x + 2y = 150

$\begin{array}{ccc}x& 50& 0\\ y& 0& 75\end{array}$

Putting (0, 0) in the inequality 3x + 2y $\le$150, we have

$3\times 0+2\times 0\le 150\Rightarrow 0\le 150$ (which is true)

So, the half plane is towards the origin.

Since, x, y $\le$ 0

So, the feasible region lies in the first quadrant.

On solving equations 4x + y = 80 and x + 5y = 115, we get A(15, 20).

Similarly, solving the equations, 3x + 2y = 150 and x + 5y = 115, we get B(40, 15).

$\therefore$ Feasible region is ABCA.

The corner points of the feasible region are A(15, 20), B(40, 15) and C(2, 72). The values of Z at these points are as follows:

$\begin{array}{cc}Cornerpoint& Z=6x+3y\\ A(15,20)& 150\to Minimum\\ B(40,15)& 285\to Maximum\\ C(2,72)& 228\end{array}$

Thus, the maximum value of Z is 285 at B(40,15).

Therefore, to maximize the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used. The maximum amount of vitamin A in the diet is 285 units.