# A car dealer determines that if gasoline-electric hybrid automobiles are sold for x dollars...

## Question:

A car dealer determines that if gasoline-electric hybrid automobiles are sold for x dollars apiece and the price of gasoline is y cents per gallon, then approximately h hybrid cars will be sold each year, where

{eq}h(x,y) = 3,500 - 19x^\frac{1}{2} + 6 (0.1y + 16)^\frac{3}{2} {/eq}

She estimates that t years from now, the hybrid cars will be selling for

x(t) = 35,050 + 350 t

dollars apiece and that gasoline will cost

{eq}y(t) = 300 + 10(3t)^\frac{1}{2} {/eq}

cents per gallon. At what rate will the annual demand for the hybrid cars be changing with respect to time 3 years from now? Will it be increasing or decreasing?

## Differentiation:

A mathematical quantity that represents the break the equation or function in the infinitesimal form to study the equation at every possible point is known as differentiation. It widely used in physics discipline.

## Answer and Explanation:

**Given Data**

- The equation for {eq}h {/eq} hybrid car sold each year is: {eq}h\left( {x,y} \right) = 3500 - 19{x^{\dfrac{1}{2}}} + 6{\left( {0.1y + 16} \right)^{\dfrac{3}{2}}} \cdots\cdots\rm{(I)} {/eq}

- The equation for hybrid car sold {eq}t {/eq} year from now a piece is: {eq}x\left( t \right) = 35050 + 350t \cdots\cdots\rm{(II)} {/eq}

- The equation for gasoline cost is: {eq}y\left( t \right) = 300 + 10{\left( {3t} \right)^{\dfrac{1}{2}}} \cdots\cdots\rm{(III)} {/eq}

- The time period from now is: {eq}t = 3\;{\rm{years}} {/eq}

Differentiate the expression (I) with respect to x

{eq}\begin{align*} \dfrac{{d\left( {h\left( {x,y} \right)} \right)}}{{dx}} &= \dfrac{{d\left( {3500 - 19{x^{\dfrac{1}{2}}} + 6{{\left( {0.1y + 16} \right)}^{\dfrac{3}{2}}}} \right)}}{{dx}}\\ &= - \dfrac{1}{2}19{\left( x \right)^{ - \dfrac{1}{2}}} \end{align*} {/eq}

Differentiate the expression (I) with respect to y

{eq}\begin{align*} \dfrac{{d\left( {h\left( {x,y} \right)} \right)}}{{dy}} &= \dfrac{{d\left( {3500 - 19{x^{\dfrac{1}{2}}} + 6{{\left( {0.1y + 16} \right)}^{\dfrac{3}{2}}}} \right)}}{{dy}}\\ &= \dfrac{3}{2}6\left( {0.1} \right){\left( {0.1y + 16} \right)^{\dfrac{1}{2}}}\\ &= 0.9{\left( {0.1y + 16} \right)^{\dfrac{1}{2}}} \end{align*} {/eq}

Differentiate the expression (II) with respect to t

{eq}\begin{align*} \dfrac{{d\left( {x\left( t \right)} \right)}}{{dt}} &= \dfrac{{d\left( {35050 + 350t} \right)}}{{dt}}\\ &= 350 \end{align*} {/eq}

Differentiate the expression (III) with respect to {eq}t {/eq}

{eq}\begin{align*} \dfrac{{d\left( {y\left( t \right)} \right)}}{{dt}} &= \dfrac{{d\left( {300 + 10{{\left( {3t} \right)}^{\dfrac{1}{2}}}} \right)}}{{dt}}\\ &= \dfrac{{10}}{2}{\left( {3t} \right)^{ - \dfrac{1}{2}}} \end{align*} {/eq}

Substitute the value and solve expression (II)

{eq}\begin{align*} x\left( 3 \right) &= 35050 + 350\left( 3 \right)\\ &= 36100 \end{align*} {/eq}

Substitute the value and solve expression (III)

{eq}\begin{align*} y\left( 3 \right) &= 300 + 10{\left( {3\left( 3 \right)} \right)^{\dfrac{3}{2}}}\\ &= 300 + 10\left( {27} \right)\\ &= 570 \end{align*} {/eq}

The expression for rate of change of annual demand is

{eq}\dfrac{{dH}}{{dt}} = \dfrac{{dH}}{{dx}}\left( {\dfrac{{dx}}{{dt}}} \right) + \dfrac{{dH}}{{dy}}\left( {\dfrac{{dy}}{{dt}}} \right) {/eq}

Substitute the value and solve the above expression

{eq}\begin{align*} \dfrac{{dH}}{{dt}} &= - \dfrac{1}{2}19{\left( x \right)^{ - \dfrac{1}{2}}}\left( x \right) + 0.9{\left( {0.1y + 16} \right)^{\dfrac{1}{2}}}\left( {\dfrac{{10}}{2}{{\left( {3t} \right)}^{ - \dfrac{1}{2}}}} \right)\\ &= - \dfrac{1}{2}19{\left( {36100} \right)^{ - \dfrac{1}{2}}}\left( {350} \right) + 0.9{\left( {0.1\left( {570} \right) + 16} \right)^{\dfrac{1}{2}}}\left( {\dfrac{{10}}{2}{{\left( {3\left( 3 \right)} \right)}^{ - \dfrac{1}{2}}}} \right)\\ &= - 17.5 + 12.81\\ &= - 4.69 \end{align*} {/eq}

Thus the rate at which annual demand for the hybrid cars be changing with respect to time 3 years from now is {eq}4.69 {/eq} per year and it is decreasing.