Answer to Question #203883 in Calculus for Bahawal tahir

Question

(a) Find fx(x,y), fy(x,y), fx(1,3), and fy(-2,4) for the given function. If 

𝑧 = 𝑓(𝑥, 𝑦) = 3𝑥

ଷ𝑦

ଶ − 𝑥

ଶ𝑦

ଷ + 4𝑥 + 9

(b) A firm estimates that it can sell Q units of its product with an advertising 

expenditure of x thousand dollars where 

𝑄 = 𝑄(𝑥) = −𝑥

ଶ + 600𝑥 + 25

i) Over what level of advertising expenditure is the number of units of 

product sold increasing? 

ii) Over what level of advertising expenditure is the number of units of 

product sold decreasing? 

c) Determine the location and values of the absolute maximum and absolute 

minimum for the given function: 

𝑓(𝑥) = (−𝑥 + 2)

ସ

, 𝑤ℎ𝑒𝑟𝑒 0 ≤ 𝑥 ≤ 3

Answer 1

(a)

$$z=f(x,y)=3x^3y^2-x^2y^3+4x+9$$

$$f_x(x, y)=9x^2y^2-2xy^3+4$$

$$f_y(x, y)=6x^3y-3x^2y^2$$

$$f_x(1,3)=9(1)^2(3)^2-2(1)(3)^3+4=31$$

$$f_y(-2,4)=9(-2)^2(4)^2-2(-2)(4)^3=832$$

(b)

$$Q(x)=-x^2+600x+25, x>0$$

$$Q'(x)=-2x+600$$

$$Q'(x)=0=>-2x+600=0$$

$$x=300$$

If $0<x<300, Q'(x)>0, Q(x)$ increases.

If $x>300, Q'(x)<0, Q(x)$ decreases.

i) The number of units of product sold is increasing for  $0<x<300.$

ii) The number of units of product sold is decreasing for  $x>300.$

(c)

$$f(x)=(-x+2)^3, 0\leq x\leq3$$

$$f'(x)=-3(-x+2)^2$$

$$f'()=0=>-3(-x+2)^2=0$$

$$x=2$$

$$f(0)=(-0+2)^3=8$$

$$f(3)=(-3+2)^3=-1$$

The function $f(x)$ has the absolute maximum with value of $8$ on $[0, 3]$ at $x=0.$

The function $f(x)$ has the absolute minimum with value of $-1$ on $[0, 3]$ at $x=3.$