16. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is ?.5 and that from a shade is ?.3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximize his profit?

0 3 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    Let the cottage industry manufacture x pedestal lamps and y wooden shades. Therefore,

    x ≥ 0 and y ≥ 0

    The given information can be compiled in a table as follows.

     

    Lamps

    Shades

    Availability

    Grinding/Cutting Machine (h)

    2

    1

    12

    Sprayer (h)

    3

    2

    20

    The profit on a lamp is Rs 5 and on the shades is Rs 3. Therefore, the constraints are

    2x + y  12

    3x + 2y  20

    Total profit, Z = 5x + 3y

    The mathematical formulation of the given problem is

    Maximize Z=5x +3y (1)

    subject to the constraints,

    2x + y  12  (2)

    3x + 2y  20  (3)

    x, y  0  (4)

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

    The corner points are A (6, 0), B (4, 4), and C (0, 10).

    The values of Z at these corner points are as follows

    The maximum value of Z is 32 at (4, 4).

    Thus, the manufacturer should produce 4 p

    ...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