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Ilia_Sergeevich [38]

3 years ago

8

I will mark as brainliest if you can answer. :)

1 answer:

Yuri [45]3 years ago

8 0

not enough information

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For the cost function c equals 0.1 q squared plus 2.1 q plus 8, how fast does c change with respect to q when q equals 11? Det

Soloha48 [4]

Answer:

Rate of change of c with respect to q is 4.3

Percentage rate of change c with respect to q is 9.95%

Step-by-step explanation:

Cost function is given as, $c=0.1\:q^{2}+2.1\:q+8$

Given that c changes with respect to q that is, $\dfrac{dc}{dq}$ . So differentiating given function,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left (0.1\:q^{2}+2.1\:q+8 \right)$

Applying sum rule of derivative,

$\dfrac{dc}{dq}=\dfrac{d}{dq}\left(0.1\:q^{2}\right)+\dfrac{d}{dq}\left(2.1\:q\right)+\dfrac{d}{dq}\left(8\right)$

Applying power rule and constant rule of derivative,

$\dfrac{dc}{dt}=0.1\left(2\:q^{2-1}\right)+2.1\left(1\right)+0$

$\dfrac{dc}{dt}=0.1\left(2\:q\right)+2.1$

$\dfrac{dc}{dt}=0.2\left(q\right)+2.1$

Substituting the value of $q=11$ ,

$\dfrac{dc}{dt}=0.2\left(11\right)+21.$

$\dfrac{dc}{dt}=2.2+2.1$

$\dfrac{dc}{dt}=4.3$

Rate of change of c with respect to q is 4.3

Formula for percentage rate of change is given as,

$Percentage\:rate\:of\:change=\dfrac{Q'\left(x\right)}{Q\left(x\right)}\times 100$

Rewriting in terms of cost C,

$Percentage\:rate\:of\:change=\dfrac{C'\left(q\right)}{C\left(q\right)}\times 100$

Calculating value of $C\left(q \right)$

$C\left(q\right)=0.1\:q^{2}+2.1\:q+8$

Substituting the value of $q=11$ ,

$C\left(q\right)=0.1\left(11\right)^{2}+2.1\left(11\right)+8$

$C\left(q\right)=0.1\left(121\right)+23.1+8$

$C\left(q\right)=12.1+23.1+8$

$C\left(q\right)=43.2$

Now using the formula for percentage,

$Percentage\:rate\:of\:change=\dfrac{4.3}{43.2}\times 100$

$Percentage\:rate\:of\:change=0.0995\times 100$

$Percentage\:rate\:of\:change=9.95%$

Percentage rate of change of c with respect to q is 9.95%

7 0

3 years ago