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

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40

7.25

step-by-step explanation:

we know that 7 equals to 7.0. we also know that 1/4=0.25. so we can add the two together and we get 7.25