A square-based, box-shaped shipping crate is designed to have a volume of 16 cubic ft. The material used to make the base costs twice as much (per square foot) as the material in the sides, and the material used to make the top costs half as much (per square foot) as the material in the sides. What are the dimensions of the crate that minimize the cost of materials

Answer:

Step-by-step explanation:

dimension

base=x, height =y, as they both are length so x, y >0

Volume of the crate= [tex]x^2y[/tex] = [tex]16ft^3[/tex]

cost

base=b unit money /[tex]ft^2[/tex]

side material = b/2 unit money /[tex]ft^2[/tex]

Top view= b/4 unit money /[tex]ft^2[/tex]

Total Cost=[tex]bx^2 + \frac{b}{2} *4xy + \frac{b}{4} *x^2[/tex] = [tex]2b( \frac{5}{8} *x^2 + xy)[/tex]

function to be minimized  = [tex]\frac{5}{8}*x^2 + xy\\[/tex]  , [tex]x^2y =16[/tex]

put [tex]y=16/x^2[/tex]

f(x) = [tex]\frac{5}{8}*x^2 + x*\frac{16}{x^2}[/tex]

upon differentiating this equation

f'(x)=0

[tex]\frac{5}{4}*x -16x^-2 = 0[/tex]

upon solving this we get [tex]x=4/\sqrt[3]{5}[/tex] [tex]y= \sqrt[3]{25}[/tex]

and further testing we get f"([tex]\frac{4}{\sqrt[3]{5} }[/tex]) >0 which proves that the point is local minimum.

Hence these are the dimensions of crate that minimizes the cost of material.

The dimensions that minimize the total cost of making the surface materials are [tex]2.3ft \text{ by } 2.3ft \text{ by }2.9ft[/tex]

[tex]T=\text{Total cost of materials}\\m_b=\text{unit cost of materials used to construct the base}\\m_t=\text{unit cost of materials used to construct the top}\\m_s=\text{unit cost of materials used to construct the sides}[/tex]

from the question, the crate has a square base and the volume is [tex]16ft^3[/tex]. So,

[tex]V=l^2h=16\\\implies h=\frac{16}{l^2}[/tex]

Also,

[tex]m_b=2m_s\\m_t=\frac{m_s}{2}[/tex]

Construct the formula for [tex]T[/tex],

[tex]T=l^2\times m_b + l^2 \times m_t+4lh\times m_s[/tex]

[tex]\text{where } l=\text{length of side of base/top}\\h=\text{height}[/tex]

Eliminating [tex]h, m_b, \text{ and }m_t[/tex], we have

[tex]T=m_s(\frac{5l^2}{2} + \frac{64}{l})[/tex]

differentiating wrt [tex]l[/tex] ([tex]m_s[/tex] is a constant), we have

[tex]\frac{dT}{dl}=m_s(5l-\frac{64}{l^2})[/tex]

The minimum total cost occurs when

[tex]m_s(5l-\frac{64}{l^2})=0[/tex]

or, when

[tex]l=\frac{4}{\sqrt[3]{5}} \approx 2.3[/tex]

(On carrying out the second derivative test, [tex]\frac{d^2T}{dl^2}>0[/tex] when [tex]l=\frac{4}{\sqrt[3]{5}}[/tex])

[tex]h=\frac{16}{l^2}=\sqrt[3]{5}^2\approx 2.9[/tex]

Therefore, the dimensions that minimize the total cost of making the surface materials are [tex]2.3ft \text{ by } 2.3ft \text{ by }2.9ft[/tex]

Another solved minimization problem can be found in the link: https://brainly.com/question/17116786

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