A manufacturer of handcrafted wine racks has determined that the cost to produce x units per month is given by upper c equals 0.2 x squared plus 10 comma 000.how fast is the cost per month changing when production is changing at the rate of 13units per month and the production level is 70​units

Respuesta :

Answer:

The cost per month is increasing at a rate $365.

Explanation:

Differentiation Formula

  • [tex]\frac{d}{dx}(x^n)= nx^{n-1}[/tex]  
  • [tex]\frac{d}{dx}(a)=0[/tex]             [ where a is a constant]
  • [tex]\frac{d}{dx}(ax^n)=a \frac{d}{dx}(x^n)= anx^{n-1}[/tex]

Given that,

A manufacturer of handcrafted wine racks has determined that the cost to produce x units per month is given by

[tex]c=0.2x^2+10,000[/tex].

Again given that,

the rate of changing production is 13 unit per month

i.e [tex]\frac{dx}{dt}=13[/tex]

To find the cost per month, we need to find out the value [tex]\frac{dc}{dt}[/tex] when production is changing at the rate 13 units per month and the production is 70 units.

[tex]c=0.2x^2+10,000[/tex]

Differentiating with respect to t

[tex]\frac{d}{dt}(c)=\frac{d}{dt}(0.2x^2)+\frac{d}{dx}(10,000)[/tex]

[tex]\Rightarrow \frac{dc}{dt}=0.2\frac{d}{dt}(x^2)+\frac{d}{dx}(10,000)[/tex]

[tex]\Rightarrow \frac{dc}{dt}=0.2\times 2x^{2-1}\frac{dx}{dt}+0[/tex]

[tex]\Rightarrow \frac{dc}{dt}=0.4x\frac{dx}{dt}[/tex]

Plugging [tex]\frac{dx}{dt}=13[/tex]

[tex]\Rightarrow \frac{dc}{dt}=0.4x\times 13[/tex]

[tex]\Rightarrow \frac{dc}{dt}=5.2x[/tex]

[tex]\frac{dc}{dt}|_{x=70}=5.2\times 70[/tex] [ plugging x=70]

            =364

[ The unit of c is not given. Assume that the unit of c is dollar.]

The cost per month is increasing at a rate $365.

ACCESS MORE