Respuesta :
Answer:
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
To minimize the construction cost, the dimensions of the cabinet should be 1.0 x [tex]\sqrt[3]{4}[/tex] x [tex]\frac{3}{2}\sqrt[3]{4}[/tex] feet.
Given the following data:
- Capacity = 2.4 [tex]ft^3[/tex].
- Cost of material (top, bottom and sides) = 90 ¢[tex]/ft^2[/tex]
- Cost of material (front and rear) = 30 ¢[tex]/ft^2[/tex]
- h = 1.5w
How to calculate the dimensions of the cup.
Let the height be h.
Let the width be w.
Let the length be L.
Therefore, the capacity (volume) of the cabinet is given by:
[tex]V = L\times w\times h\\\\2.4=Lw(1.5w)\\\\2.4=1.5Lw^2\\\\1.6=Lw^2\\\\L=\frac{1.6}{w^2}[/tex]
For the area of top and bottom:
[tex]A=2(Lw)=2Lw\\\\A=2 \times \frac{1.6}{w^2} \times w\\\\A=\frac{3.2}{w}[/tex]
For the cost of top and bottom:
[tex]C_1=90 \times \frac{3.2}{w} \\\\C_1=\frac{288}{w}[/tex]
For the area sides:
[tex]A=3(Lw)=3Lw\\\\A=3 \times \frac{1.6}{w^2} \times w\\\\A=\frac{4.8}{w}[/tex]
For the cost of sides:
[tex]C_2=90 \times \frac{4.8}{w} \\\\C_2=\frac{432}{w}[/tex]
For the area of front and rear:
[tex]A=2(hw)=2hw\\\\A=2 \times 1.5w \times w\\\\A=3.2w^2[/tex]
For the cost of front and rear:
[tex]C_3=30 \times 3w^2 \\\\C_3=90w^2[/tex]
At the minimum cost, we have:
[tex]C(w)=C_1+C_2+C_3\\\\C(w)=\frac{288}{w} +\frac{432}{w} +90w^2[/tex]
Differentiating wrt w, we have:
[tex]C(w)'=-\frac{288}{w^2} -\frac{432}{w^2} +180w=0\\\\-\frac{288}{w^2} -\frac{432}{w^2} +180w=0\\\\180w=\frac{288}{w^2}+\frac{432}{w^2}\\\\180w^3=288+432\\\\180w^3=720\\\\w^3=4\\\\w=\sqrt[3]{4} \;ft[/tex]
For the height:
[tex]h=1.5w=\frac{3}{2} w\\\\h=\frac{3}{2}\sqrt[3]{4} \;ft[/tex]
For the length:
[tex]L=\frac{1.6}{w^2} =\frac{\frac{8}{5} }{w^2} \\\\L=\frac{8}{5w^2} \\\\L=\frac{8}{5(\sqrt[3]{4} )^2}[/tex]
L = 1.0 ft.
Read more on capacity here: brainly.com/question/25248189
