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scoundrel [369]
3 years ago
6

A polynomial that has 5 turning points is multiplied by another polynomial that has 6 zeros. What is the minimum degree of the n

ew polynomial?
Mathematics
1 answer:
Rom4ik [11]3 years ago
8 0
The only thing in the question I am not totally certain about is the definition of a 'zero' on a polynomial - I will assume this means an x-intercept, which seems to make sense since these are the points where the value of the function is zero.

Taking the first polynomial: the maximum number of turning points for a polynomial of order n is (n-1). Take the example of a quadratic, which always has 1 turning point. Therefore the minimum order of the first polynomial is 6.

The maximum number of x-intercepts for a polynomial of order n is n. Therefore the second polynomial has a minimum order of 6.

Multiplication of two polynomials can get very messy very quickly. However, picture putting the two in brackets next to each other, such that the terms are in decreasing orders. You can easily see the maximum order term is found by multiplying the first term in each bracket. A polynomial's order is judged solely on the maximum power of the variable, so this is all we need to consider.

This multiplication becomes ax^6 * bx^6 where a and b are arbitrary constants in this context. Hence this product is abx^12 (exponents add when the terms are multiplied, where 12 = 6 + 6), so minimum degree of new polynomial = 12
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Bob makes three different silverware - spoons, forks and knives. Each unit (dozen) of spoon gives him a profit of $9, forks of $
Gala2k [10]

Answer:

Solution is:

z  (max) = 687.5

x₁   =  0

x₂   = 16

x₃   = 21

Step-by-step explanation:

From the problem statement we have:

Resources:      machine time ( 183 h)      labor (250  h)   steel  185 (pounds)

Unit

Spoons                          4                                     4                          3

forks                               4                                     9                          2

knives5                          5                                     5                          4

Profit $                           9                                    20                       17.5

Objective function z =  9*x₁  +  20*x₂  + 17.5 *x₃   to maximize

Subject to:

Availability of machine time  :  183 h

4*x₁  +  4*x₂ +   5*x₃ ≤  183

Availability of labor  :  250 h

4*x₁  +  9*x₂  +  5*x₃  ≤  250

Availability of steel  :  185 pounds

3*x₁  +  2*x₂  +  4*x₃  ≤  185

Requirement:

x₂ ≥ 16

General constraints:

x₁  ≥ 0             x₃      ≥  0 all integers

After 6 iteration the solution using AtomZmath on-line solver

z  (max) = 687.5

x₁   =  0

x₂   = 16

x₃   = 21

Resources used:

Machine time: 16* 4 + 21*5  =  64  +  105  = 169

remains   183 - 169  = 14 h

Labor:  16*9  +  21* 5  =  144  +  105  =  249

remains   250 -  249  =  1 h

Steel :  16*2  +  21*4  =  32  + 84  =  116

remains   185  -  116  = 69 pounds.

If it is decided that 20 units of forks are to be made then

we will need   4*4 = 16   h of machine time

                         9*4 =  36 h  of labor

                         2*4 = 8 pounds of steel

We can get that from abandom to make one unit of x₃ ??

No because as we said we need 36 hours  of  labor ( we still have 1 we need 35 more hours ) if we make 20 x₃ insted of 21 we get only 5 hours.

z we got is maximum

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3 years ago
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