MATHEMATICS COLLEGE

A Jewelry store makes necklaces and bracelets from gold and platinum. vThe store has 18 ounces of gold and 20 ounces of platinum. vEach necklace requires 3 ounces of gold and 2 ounces of platinum, whereas each bracelet requires 2 ounces of gold and 4 ounces of platinum. vThe demand for bracelets is no more than four. vA necklace earns $300 in profit and a bracelet $400. vThe store wants to determine the number of necklaces and bracelets to produce in order to maximize profit.a. Formulate a linear programming model for this problem.b. Solve this model using graphical analysis.

Answers

Answer 1

Answer:

Answer:

maximum profit is$2400 when 4 necklace and 3 brackets are made.

Step-by-step explanation:

Total gold = 18 ounces

Total platinum = 20 ounces.

let X₁ represents the necklace and X₂ represents the bracelets.

A. Linear Programming Model

maximize:

$300x_(1) + 400x_(2)$

with constraints:

for gold:

$3x_(1) + 2x_(2) \leq 18$ ---(1)

for platinum:

$2x_(1) + 4x_(2) \leq 20$ ---(2)

The demand for bracelets is no more than four i.e.

$x_(2)\leq 4$ ---(3)

$x_(1),x_(2)\geq 0$

B. Graphical Analysis

Answer 2

Answer:

Final answer:

To maximize profit, formulate a linear programming model with constraints for the number of necklaces and bracelets to produce. Solve the model using graphical analysis to find the optimal solution.

Explanation:

To formulate a linear programming model for this problem, let x be the number of necklaces to produce and y be the number of bracelets to produce. The objective is to maximize profit, which can be expressed as: Profit = 300x + 400y. The constraints are: 3x + 2y ≤ 18 (gold constraint), 2x + 4y ≤ 20 (platinum constraint), 0 ≤ x ≤ infinity (non-negativity constraint), and y ≤ 4 (demand constraint).

To solve this model using graphical analysis, graph the feasible region determined by the constraints. The feasible region is the region in which all constraints are satisfied. The optimal solution will be at one of the corner points of the feasible region. Calculate the objective function at each corner point and select the one that maximizes profit.

Learn more about Linear Programming here:

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Answers

Answer:

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Step-by-step explanation:

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Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. ln(x^2+1)

Answers

Answer:

The minimum value of the given function is f(0) = 0

Step-by-step explanation:

Explanation:-

Extreme value :- f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.

i) the necessary and sufficient condition for f(x) to have a maximum or minimum at given point.

ii) find first derivative $f^(l) (x)$ and equating zero

iii) solve and find 'x' values

iv) Find second derivative $f^(ll)(x) >0$ then find the minimum value at x=a

v) Find second derivative $f^(ll)(x) <0$ then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

step1:- find first derivative $f^(l) (x)$ and equating zero

$f^(l)(x) = (1)/(x^2+1) (d)/(dx)(x^2+1)$

$f^(l)(x) = (1)/(x^2+1) (2x)$ ……………(1)

$f^(l)(x) = (1)/(x^2+1) (2x)=0$

the point is x=0

step2:-

Again differentiating with respective to 'x', we get

$f^(ll)(x)=(x^2+1(2)-2x(2x))/((x^2+1)^2)$

on simplification , we get

$f^(ll)(x) = (-2x^2+2)/((x^2+1)^2)$

put x= 0 we get $f^(ll)(0) = (2)/((1)^2)$ > 0

$f^(ll)(x) >0$ then find the minimum value at x=0

Final answer:-

The minimum value of the given function is f(0) = 0

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Answer:

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Step-by-step explanation:

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