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Chapter 5: Problem 1

Find the maximum value of the objective function $$ z=2 x^{2}-3 x y+2 y+10 $$ subject to the constraint \(y=x\).

Short Answer

Expert verified
The maximum value of \(z\) is 11.

Step by step solution

01

Substitute the Constraint

Given the constraint is \(y = x\), substitute \(y\) with \(x\) in the objective function. The objective function becomes: \[ z = 2x^2 - 3x(x) + 2(x) + 10 \] which simplifies to \[ z = 2x^2 - 3x^2 + 2x + 10 = -x^2 + 2x + 10 \]
02

Simplify the Objective Function

The new objective function is: \[ z = -x^2 + 2x + 10 \] which is a quadratic function. Identify the coefficients: \(a = -1\), \(b = 2\), and \(c = 10\).
03

Find the Vertex of the Parabola

The maximum or minimum value of a quadratic function \(ax^2 + bx + c\) occurs at the vertex, given by \(-\frac{b}{2a}\). Here, \(a = -1\) and \(b = 2\), so: \[ x = -\frac{2}{2(-1)} = 1 \]
04

Calculate the Maximum Value

Substitute \(x = 1\) back into the simplified objective function to find \(z\). \[ z = -1^2 + 2(1) + 10 = -1 + 2 + 10 = 11 \] Thus, the maximum value of \(z\) is 11.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Objective Function
The objective function is the main equation in an optimization problem that you aim to maximize or minimize. In this exercise, the given objective function is: \[ z = 2x^2 - 3xy + 2y + 10\]Our goal is to find the maximum value of this function, considering the given constraint. Understanding the objective function's role is crucial because it represents the quantity we want to optimize. It often involves multiple variables.
Constraint Substitution
Constraint substitution simplifies optimization problems by reducing the number of variables. Here, the constraint is given as: \[ y = x\]We substitute \( y \) with \( x \) in the objective function. This changes the function to: \[ z = 2x^2 - 3x(x) + 2(x) + 10 = -x^2 + 2x + 10\]This transformation simplifies the problem, converting it into a single-variable function. This step is essential for making the optimization process manageable.
Quadratic Function
A quadratic function is a polynomial of degree 2, generally written as: \[ ax^2 + bx + c\]In this problem, after substituting the constraint, the simplified objective function is: \[ z = -x^2 + 2x + 10\]Here, \( a = -1 \), \( b = 2 \), and \( c = 10 \). Quadratic functions graph as parabolas, and the sign of \( a \) determines whether the parabola opens upwards (\( a > 0 \)) or downwards (\( a < 0 \)). Since \( a = -1 \), the parabola opens downwards, indicating the vertex will give the maximum value of the function.
Vertex of Parabola
The vertex of a parabola represents the peak (maximum) or the trough (minimum) of the quadratic function. For the quadratic function \( ax^2 + bx + c \), the vertex occurs at: \[ x = -\frac{b}{2a}\]Using the coefficients from our function \( z = -x^2 + 2x + 10 \), we find the vertex by plugging in \( a = -1 \) and \( b = 2 \): \[ x = -\frac{2}{2(-1)} = 1\]Substitute \( x = 1 \) back into the function to find the maximum \( z \): \[ z = -1^2 + 2(1) + 10 = -1 + 2 + 10 = 11\]Thus, the vertex is at \( x = 1 \) and the maximum value of \( z \) is 11.

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Most popular questions from this chapter

(Maple) A monopolistic producer charges different prices at home and abroad. The demand functions of the domestic and foreign markets are given by $$ P_{1}+Q_{1}=100 \quad \text { and } \quad P_{2}+2 Q_{2}=80 $$ respectively. The firm's total cost function is $$ \mathrm{TC}=\left(Q_{1}+Q_{2}\right)^{2} $$ (a) Show that the firm's profit function is given by $$ \pi=100 Q_{1}+80 Q_{2}-2 Q_{1}^{2}-3 Q_{2}^{2}-2 Q_{1} Q_{2} $$ Use calculus to show that profit is maximized when \(Q_{1}=22\) and \(Q_{2}=6\), and find the corresponding prices. (b) The foreign country believes that the firm is guilty of dumping because the good sells at a higher price in the home market, so decides to restrict the sales to a maximum of 2 , so that \(Q_{2} \leq 2\). By plotting \(\pi\) in the region \(0 \leq Q_{1} \leq 30,0 \leq Q_{2} \leq 2\), explain why the profit is maximized when \(Q_{2}=2\). Use calculus to find value of \(Q_{1}\), and compare the corresponding profit with that of the free market in part (a).

Find expressions for the partial derivatives \(f_{1}, f_{11}\) and \(f_{21}\) in the case when $$ f\left(x_{1}, x_{2}, x_{3}\right)=x_{1} x_{2}+x_{1}^{5}-x_{2}^{2} x_{3} $$

Given the demand function $$ Q=500-3 P-2 P_{\mathrm{A}}+0.01 Y $$ where \(P=20, P_{A}=30\) and \(Y=5000\), find (a) the price elasticity of demand (b) the cross-price elasticity of demand (c) the income elasticity of demand If income rises by \(5 \%\), calculate the corresponding percentage change in demand. Is the good inferior or superior?

Consider the four-sector model $$ \begin{array}{ll} Y=C+I+G+X-M & \\ C=a Y+b & (00) \\ I=I^{*} & \left(I^{*}>0\right) \\ G=G^{*} & \left(G^{*}>0\right) \\ X=X^{*} & \left(X^{*}>0\right) \\ M=m Y+M^{*} & \left(00\right) \end{array} $$ where \(X\) and \(M\) denote exports and imports respectively and \(m\) is the marginal propensity to import. (a) Show that $$ Y=\frac{b+I^{*}+G^{*}+X^{*}-M^{*}}{1-a+m} $$ (b) Write down the autonomous export multiplier $$ \frac{\partial Y}{\partial X^{*}} $$ and the marginal propensity to import multiplier $$ \frac{\partial Y}{\partial m} $$ Deduce the direction of change in \(Y\) due to increases in \(X^{*}\) and \(m\). (c) If \(a=0.8, b=120, l^{*}=100, G^{*}=300, X^{*}=150, m=0.1\) and \(M^{*}=40\), calculate the equilibrium level of national income, \(Y\), and the change in \(Y\) due to a 10 unit increase in autonomous exports.

(Maple) Consider the production function $$ Q=\left(0.3 K^{-3}+0.7 L^{-3}\right)^{-1 / 3} \quad(1 \leq K \leq 10,1 \leq L \leq 10) $$ (a) Draw a three-dimensional plot of this function. Rotate the axes to give a clear view of the surface. (b) Draw the corresponding isoquant map. Deduce that the marginal rate of technical substitution diminishes with increasing \(L\). (c) Find an expression for MRTS. (d) Find the slope of the isoquant \(Q=4\) at the point \(L=8\).
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