18. A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs. 25000 and Rs. 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than ? 70 lakhs and if his profit on the desktop model is ?. 4500 and on portable model is ?. 5000.

0 2 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    Let the merchant stock x desktop models and y portable models. Therefore,

    x ≥ 0 and y ≥ 0

    The cost of a desktop model is Rs 25000 and of a portable model is Rs 4000. However, the merchant can invest a maximum of Rs 70 lakhs.

    25000x + 40000y  7000000

    5x +8y  1400

    The monthly demand of computers will not exceed 250 units.

    x+y250

    The profit on a desktop model is Rs 4500 and the profit on a portable model is Rs 5000.

    Total profit, Z=4500x +5000y

    Thus, the mathematical formulation of the given problem is

    Maximum Z=4500x+5000y             .....(1)

    subject to the constraints,

    5x +8y  1400        ....(2)

    x + y  250       .....(3)

    x, y  01400        ......(4)

    The feasible region determined by the system of constraints is as follows.

    The corner points are A (250, 0), B (200, 50

    ...more

Similar Questions for you

A
alok kumar singh

Δ | 3 2 k 2 4 2 1 2 1 | = | 3 2 k 0 8 0 1 2 1 | , [ R 2 R 1 2 R 3 ]

= -8 (-3 + k)

For inconsistent Δ = 0 k = 3  

Δ x = | 1 0 2 k 6 4 2 5 m 2 1 | = 3 2 4 0 m 0 m 4 5              

V
Vishal Baghel

  l = π 2 π 2 ( [ x ] + [ s i n x ] ) d x . (i)

I = π 2 π 2 ( [ x ] + [ s i n x ] ) d x . (ii)

by using property ( a b f ( x ) d x = a b f ( a + b x ) d x )

Adding (i) and (ii) we get 2l = π 2 π 2 ( 2 ) d x = 2 π l = π

A
alok kumar singh

l n , m = 0 1 2 x n x m 1 d x , m , n N , n > m

l 6 + i , 3 l 3 + i , 3

A = 1 2 5 [ 1 5 1 5 1 5 0 1 1 2 1 1 2 0 0 1 2 8 ] = B 3 2

| A | = ( 1 3 2 ) 3 | B | = 1 1 0 5 . 2 1 9

V
Vishal Baghel

α + β + γ = 2 π

Δ = | 1 c o s γ c o s β c o s γ 1 c o s α c o s β c o s α 1 |

= 1 c o s 2 α c o s 2 β c o s 2 γ + 2 c o s α . c o s β . c o s γ

= c o s γ c o s ( α β ) + c o s γ c o s ( α β ) = 0

V
Vishal Baghel

Given 2x + y – z = 3         . (i)

x – y – z = α        . (ii)

3x + 3y + βz = 3                . (iii)

(i) x 2 – (ii) – (iii) – (1 + β) z = 3 - α

For infinite solution 1 + β = 0 = 3 - α

=> α = 3, β = -1

So, α + β - αβ = 5

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 687k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post