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The output of a certain plant is Q = 0.07x^2 + 0.14xy + 0.05y^2 units per day, where x is the...

Question:

The output of a certain plant is {eq}Q = 0.07x^2 + 0.14xy + 0.05y^2 {/eq} units per day, where {eq}x {/eq} is the number of hours of skilled labour used and {eq}y {/eq} is the number of hours of unskilled labour used. Currently, 50 hours of skilled labour and 600 hours of unskilled labour are used each day.

Use calculus to estimate the change in unskilled labour that should be made to offset a 1-hour increase in skilled labour so that output will be maintained at its current level.

Solving Quadratic Equations, Implicit Differentiation:

A word problem describing a real time situation of a quantity dependent on one or more variables can be formulated oftentimes into a simple algebraic equation. The equation would normally represent a polynomial. Given certain initial conditions, such a polynomial equation can be reduced to a polynomial in a single variable of varying degrees. A second degree polynomial in a single variable equated to zero is called a quadratic equation. The quadratic equation can have two identical roots, or two distinct real roots or two imaginary roots.

For example, the quadratic equation, {eq}x^2 + 1 = 0 {/eq} has no real roots.

The equation, {eq}(x+1)^2 = 0 {/eq} has two identical roots, both equal to {eq}x = -1 {/eq}

The formula to find the roots of a quadratic equation of the form {eq}ax^2 + bx + c = 0 {/eq} is,

{eq}\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} {/eq}.

We can also use the method of implicit differentiation when we wish to find the derivative {eq}\frac{dy}{dx} {/eq} when the dependent variable {eq}y {/eq} is not explicitly expressed in terms of {eq}x {/eq}, the dependent variable. We use the chain rule in differentiation in this case.

The chain rule is given by,

{eq}\displaystyle \frac{d(Q(y))}{dx} = \frac{dQ}{dy}\frac{dy}{dx} {/eq}

Answer and Explanation:

We can solve this problem two ways, using the quadratic equation or calculus.

First, we will demonstrate how it can be solved using algebra and the quadratic equation.

We are given that the function representing the output of a certain plant is {eq}Q = 0.07x^2 + 0.14xy + 0.05y^2 {/eq} units per day, where {eq}x {/eq} is the number of hours of skilled labor used and {eq}y {/eq} is the number of hours of unskilled labor used.

We are informed that presently 50 hours of skilled labor and 600 hours of unskilled labor are used each day. Therefore the current total output per day would be,

{eq}\begin{align*} Q(50, 600) &= 0.07 \times 50^2 + 0.14(50)(600) + 0.05(600^2) \\ &= 0.07 \times 2500 + 0.14 \times 30000 + 0.05 \times 360000 \\ &= 175 + 4200 + 18000 \\ &= 4375 +18000 \\ &= 22375\ units\ per\ day \end{align*} {/eq}

We are required to estimate the change in unskilled labor that should be made to offset a 1-hour increase in skilled labor so that output will be maintained at its current level.

A one hour increase in skilled labor implies that the number of hours of skilled labor to be used per day is now, {eq}x + 1 {/eq}.

We want to find the corresponding value of {eq}y {/eq} which is the number of hours of unskilled labor to be used so that the output per day is still, {eq}22375 {/eq} units per day.

Therefore we have the equation,

{eq}Q = 22375 = 0.07(x+1)^2 + 0.14(x+1)y +0.05y^2 {/eq}

We know that the current value for the number of hours of skilled labor: {eq}x = 50 {/eq}

We substitute this value in the above equation to get,

{eq}Q =22375 = 0.07(50+1)^2 + 0.14(50+1)y + 0.05y^2 {/eq}

The above equation is simply a quadratic equation in the single variable {eq}y {/eq} and can be solved using the formula for finding quadratic roots. We simplify the above equation and solve for {eq}y {/eq} as follows:

{eq}\begin{align*} 22375 &= 182.07 + 7.14y + 0.05y^2 \\ 0 &= -22375 + 182.07 + 7.14y + 0.05y^2 \\ 0 &= - 22192.93 + 7.14y + 0.05y^2 \\ y &= \frac{-7.14 \pm \sqrt{7.14^2 + 4 \times (0.05) \times 22192.93 }}{2(0.05)} \\ &= \frac{-7.14 \pm \sqrt{50.9796 + 4438.586}}{0.1} \\ &= \frac{-7.14 \pm \sqrt{4489.5656 }}{0.1} \\ &\approx \frac{-7.14 \pm 67.004}{0.1} \\ &= 598.64 , -741.44\\ y &\approx 598.64 \end{align*} {/eq}

hours per day. We choose the positive root so that we have some positive hours of unskilled labor in the plant.

Therefore the change in unskilled labor would be {eq}598.64 - 600 = -1.36 {/eq}.

The plant would have a decrease of {eq}1.36 {/eq} hour of unskilled labor per day to maintain the same output per day, for an increase in {eq}1 {/eq} hour of skilled labor per day.


Now we solve the problem using Calculus. We find the derivative {eq}\displaystyle \frac{dy}{dx} {/eq} with respect to {eq}x {/eq} using implicit differentiation.

{eq}\begin{align*} Q &= 0.07(x+1)^2 + 0.14(x+1)y + 0.05y^2 \\ \frac{dQ}{dx} &= 0.14( x + 1) + 0.14 y + 0.14( x + 1) \frac{dy}{dx} + 0.10 y \frac{dy}{dx} \\ \frac{dQ}{dx} &= 0.14( x + 1) + 0.14 y + (0.14( x + 1) + 0.10 y )\frac{dy}{dx} \\ \end{align*} {/eq}

From the above differential equation we can solve for the change in unskilled labour that should be made to offset a 1-hour increase in skilled labour so that output will be maintained at its current level.

Since the output is to remain unchanged we have the relation,

{eq}\displaystyle \frac{dQ}{dx} = 0 {/eq}.

We are given the change in skilled labor,

{eq}dx = +1 {/eq}.

We are also given the current values for skilled and unskilled labor as,

{eq}x = 50\\ y = 600 {/eq}

We must use the above values to find the corresponding change in unskilled labor, {eq}dy {/eq}.

Substituting the above values in our differential equation we get,

{eq}\begin{align*} \frac{dQ}{dx} &= 0.14 \times ( x + 1) + 0.14 y + (0.14( x + 1) + 0.10 y )\frac{dy}{dx} \\ 0 &= 0.14 \times ( 50 + 1) + 0.14 \times 600 + (0.14( 50 + 1) + 0.10 \times 600 )\frac{dy}{1} \\ dy &= - \frac{0.14 ( 51 + 600)}{0.14 \times 51 + 0.10 \times 600} \\ &= - \frac{0.14 (651)}{0.14 \times 51 + 0.10 \times 600} \\ &= - \frac{91.14}{7.14 + 60} \\ &= - \frac{91.14}{67.14} \\ dy &\approx - 1.36 \end{align*} {/eq}

{eq}\text{The change in unskilled labour that should be made to offset a 1-hour increase in skilled labour so that output will be maintained at its current level is a 1.36 - hour decrease. } {/eq}


Learn more about this topic:

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The Quadratic Formula: Definition & Example

from Remedial Algebra I

Chapter 25 / Lesson 10
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