1. A manufacturer produces two types of products - A and B, which are subjected to two types of operations, viz. grinding and polishing. Each unit of product A takes 2 hours of grinding and 3 hours of polishing whereas product B takes 3 hours of grinding and 2 hours of polishing. The manufacturer has 10 grinders and 15 polishers. Each grinder operates for 12 hours/day and each polisher 10 hours/day. The profit margin per unit of A and B are Rs. 5/ - and Rs. 7/ - respectively. If the manufacturer utilises all his resources for producing these two types of items, what is the maximum profit that the manufacturer can earn?
Write Comment
Comments
By: anil on 05 May 2019 02.38 am
Let the number of units of A and B produced be $$x$$ and $$y$$ respectively. For product A, time taken for grinding = $$2x$$ and polishing = $$3x$$ For product B, time taken for grinding = $$3y$$ and polishing = $$2y$$
Total number of hours of grinding done per day = $$10 imes 12 = 120$$ hrs Total number of hours of polishing done per day = $$15 imes 10 = 150$$ hrs
=> $$2x + 3y = 120$$ ---------Eqn(I) and $$3x+ 2y = 150$$ ----------Eqn(II) Applying the operation : 3* Eqn(I) - 2* Eqn(II), we get : => $$(6x - 6x) + (9y - 4y) = 360 - 300$$ => $$y = frac{60}{5} = 12$$ => $$x = 42$$ $$ herefore$$ Profit made by the manufacturer = $$(42 imes 5) + (12 imes 7)$$ = $$210 + 84 = Rs. 294$$
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use
Total number of hours of grinding done per day = $$10 imes 12 = 120$$ hrs Total number of hours of polishing done per day = $$15 imes 10 = 150$$ hrs
=> $$2x + 3y = 120$$ ---------Eqn(I) and $$3x+ 2y = 150$$ ----------Eqn(II) Applying the operation : 3* Eqn(I) - 2* Eqn(II), we get : => $$(6x - 6x) + (9y - 4y) = 360 - 300$$ => $$y = frac{60}{5} = 12$$ => $$x = 42$$ $$ herefore$$ Profit made by the manufacturer = $$(42 imes 5) + (12 imes 7)$$ = $$210 + 84 = Rs. 294$$