The cabinet that will enclose the acrosonic model d loudspeaker system will be rectangular and will have an internal volume of 2.4 ft3. for aesthetic reasons, the design team has decided that the height of the cabinet is to be 1.5 times its width. if the top, bottom, and sides of the cabinet are constructed of veneer costing 90¢/ft2 and the front (ignore the cutouts in the baffle) and rear are constructed of particle board costing 30¢/ft2, what are the dimensions of the enclosure that can be constructed at a minimum cost?

Dimensions of cabinet

x (wide) = 1.93 ft

y (hight) = 2.895 ft

p (depth) =0.43 ft

Step-by-step explanation:

Dimensions of cabinet

y  height

x  wide

p  deph

From problem statement

y = 1.5 x       V = y * x * p      V = 1.5*x²p      but p = V/y*x     p = 2.4/1.5 x²

p = 1.6 / x²

Then

Area of top and bottom    A₁ = 2*x*p  ⇒   2*x*1.6/x²  

A₁  = 3.2 /x

And cost  in $ C₁  = 0,9 * 3.2 /x     ⇒   C₁  = 2.88/x

Area of sides (front and rear not included)

A₂ = 2*y *p      A₂ = 3*x*1.6/x²    A₂ = 4.8/x

And cost in $ C₂ = 0.9 * 4.8 /x    C₂ = 4.32 /x

Area of  front and rear   A₃  =2* y*x    A₃  = 2*1.5 *x²    A₃ = 3x²

And cost  C₃ = 0.3 * 3/x²    =  0.9/x²

Total cost  C(x) = C₁  + C₂ + C₃            C(x) = 2.88/x  +  4.32/x + 0.9x²

Taking derivatives

C´(x)  =  -2.88/x² - 4.32 /x² + 0.9 x

C´(x) = 0             -2.88/x² - 4.32/x² + 0.9 x = 0     -2.88 - 4.32 + 0.9 x³ = 0

-7.2 + x³ = 0    x³ = 7.2    

x = 1.93 ft        y = 1.5*1.93  = 2.895 ft    and p = 0.43 ft

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.