A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs atleast 240 kg of phosphoric acid, atleast 270 kg of potash and atmost 310 kg of chlorine. If the grower wants to maximize the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?

$\begin{matrix} B r a n d P & B r a n d Q N i t r o g e n & 3 & 3.5 P h o s p h o r i c a c i d & 1 & 2 P o t a s h & 3 & 1.5 C h l o r i n e & 1.5 & 2 \end{matrix}$

Open in App

Solution

Let the fruit grower mixes x bags of brand P and y bags of brand Q. Construct the following table:

$\begin{matrix} B r a n d o f & N u m b e r o f & A m o u n t o f & A m o u n t o f & A m o u n t o f & A m o u n t o f f e r t i l i z e r & b a g s & n i t r o g e n & p h o s p h o r i c a c i d & p o t a s h & c h l o r i n e P & x & 3 x & 1 x & 3 x & 1.5 x Q & y & 3.5 y & 2 y & 1.5 y & 2 y T o t a l & x + y & 3 x + 3.5 y & x + 2 y & 3 x + 1.5 y & 1.5 x + 2 y \end{matrix}$

Our problem is to minimize

Z = 3x + 3.5y .....(i)

$S u b j e c t t o t h e c o n s t r a i n t s a r e x + 2 y \geq 240 . . (i i) e x + 1.5 y \geq 270 . . (i i i) 1.5 x + 2 y \leq 310 . . (i v) x \geq 0, y \geq 0 F i r s t l y, d r a w t h e g r a p h o f t h e l i n e x + 2 y = 240 \begin{matrix} x & 0 & 240 y & 120 & 0 \end{matrix} P u t t i n g (0, 0) i n t h e i n e q u a l i t y x + 2 y \geq 240, w e h a v e 0 + 2 \times 0 \geq 240 \Rightarrow 0 \geq 240 (w h i c h i s f a l s e) S o, t h e h a l f p l a n e i s a w a y f r o m t h e o r i g i n . S e c o n d l y, d r a w t h e g r a p h o f t h e l i n e 3 x + 1.5 y = 270 \begin{matrix} x & 0 & 90 y & 180 & 0 \end{matrix} P u t t i n g (0, 0) i n t h e i n e q u a l i t y 3 x + 1.5 y \geq 270, w e h a v e 3 \times 0 + 1.5 \times 0 \geq 270 \Rightarrow 0 \geq 270 (w h i c h i s f a l s e) S o, t h e h a l f p l a n e i s a w a y f r o m t h e o r i g i n . T h i r d l y, d r a w t h e g r a p h o f t h e l i n e 1.5 x + 2 y = 310 \begin{matrix} x & 0 & 620 / 3 y & 155 & 0 \end{matrix} P u t t i n g (0, 0) i n t h e i n e q u a l i t y 1.5 x + 2 y \leq 310, w e h a v e 1.5 \times 0 + 2 \times 0 \leq 310 \Rightarrow 0 \leq 310 (w h i c h i s t r u e) S o, t h e h a l f p l a n e i s t o w a r d s t h e o r i g i n .$

the intersection point of lines 3x+1.5y = 270 and 1.5x+2y = 310 is B(20, 140); of lines 1.5x + 2y = 310 and x + 2y =240 is A(140, 50); of lines 2 + 2y = 240 and 3x + 1.5y = 270 is C(40, 100).

Since, x,y $\geq$ 0

So, the feasible region lies in the first quadrant.

Let Z be the total cost.

The corner points of the feasible region are C(40, 100), A(140, 50) and B(20, 140). The values of Z at these points are as follows:

$\begin{matrix} C o r n e r p o i n t & Z = 3 x + 3.5 y A (140, 50) & 595 \to M i n i m u m B (20, 140) & 550 C (40, 100) & 470 \end{matrix}$

The maximum value of Z is 595 at (140, 50).

Thus, 140 bags of brand P and 50 bags of brand Q should be added to the garden to maximize the amount of nitrogen.
The maximize amount of nitrogen added to the garden is 595 kg.

Suggest Corrections

Similar questions

A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and atmost 310 kg of chlorine. If the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used. What is the minimum amount of nitrogen added in the garden?

$\begin{matrix} k g p e r b a g \begin{matrix} F e r t i l i z e r & B r a n d P & B r a n d Q N i t r o g e n & 3 & 3.5 P h o s p h o r i c a c i d & 1 & 2 P o t a s h & 3 & 1.5 C h l o r i n e & 1.5 & 2 \end{matrix} \end{matrix}$