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Question: A sweet – shop makes gift packets of sweets by combining two special types of sweets A and B which a...

A sweet – shop makes gift packets of sweets by combining two special types of sweets A and B which are 7 kg. At least 3 kg of A and no more than 5 kg of B should be used. The shop makes a profit of Rs. 15 on A and Rs. 20 on B per kg. Determine the product mix so as to obtain maximum profit.

Explanation

Solution

Hint:In this question we are given to obtain a maximum profit of product with the combination of sweets A and B. To calculate this first we need to find the profit of a product mix with different quantities by taking few points into consideration which is given in the question, i.e. sweet A should be equal or greater than 3 (A3)\left( A\ge 3 \right) and sweet B should be equal or less than 5 (B5)\left( B\le 5 \right) then compare the profit of both products as to ascertain the answer.

Complete step-by-step answer:
From the question we get to know that a gift packed by a sweet shop is a combination of two sweets such as A and B which is of 7 kg i.e.A+B=7kgA+B=7kg .......... (1)
To make this product mix we have to take at least 3 kg of A and not more than 5 kg of B i.e. A3A\ge 3 and B5B\le 5 .
It is given that a shop makes a profit of Rs 15/- on A and Rs 20/- on B per kg.
Let us consider that the product 1 contains: A=3kgA=3kg and B=4kg.B=4kg.
We know that the profit on sweet A is Rs 15/- per kg and we have 3kg. So, we can take the total profit on A as (15×3)\left( 15\times 3 \right). And the profit on sweet B is Rs 20/- per kg and we have 4kg. So, we can take the total profit on sweet B as (4×20)\left( 4\times 20 \right) .
Thus, we will get the profit earned on product 1 by taking the total profit earned on A and B in equation (1) we get –
(3×15)+(4×20)\left( 3\times 15 \right)+\left( 4\times 20 \right)
=45+80=45+80
=125.=125.
Let us consider that the product 2 contains: A=4kgA=4kg and B=3kgB=3kg
We know that the profit on sweet A is Rs 15/- per kg and we have 4kg. So, we can take the total profit on A as (15×4)\left( 15\times 4 \right). And the profit on sweet B is Rs 20/- per kg and we have 3kg. So, we can take the total profit on sweet B as (3×20)\left( 3\times 20 \right) .
Thus, we will get the profit earned on product 2 by taking the total profit earned on A and B in equation (1) we get –
(4×15)+(3×20)\left( 4\times 15 \right)+\left( 3\times 20 \right) .
= 60+60
= 120.
Therefore, the profit earned on product 1 > product 2.
Hence, the product mix of A = 3kgA\text{ }=\text{ }3kg and B = 4kgB\text{ }=\text{ }4kg can obtain maximum profit.

Note: Students can solve this problem directly on a logical basis.
For example: The shop makes a maximum profit of Rs 20/- on B than A per kg, we will take the maximum amount of B in the product mix.
As the maximum limit of B is 5 kg in a product for which we have to take A = 2 kgA\text{ }=\text{ }2\text{ }kg , that can’t be possible because, the minimum limit of A in a product mix is 3 kg.So, we will take 4 kg of B and 3 kg of A to obtain maximum profit.