Respuesta :
Answer:
The absolute minimum of the surface area function on the interval [tex](0,\infty)[/tex] is [tex]S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2[/tex]
The dimensions of the box with minimum surface area are: the base edge [tex]x=2\sqrt[3]{15}\:ft[/tex] and the height [tex]h=\sqrt[3]{15} \:ft[/tex]
Step-by-step explanation:
We are given the surface area of a box [tex]S(x)=x^2+\frac{240}{x}[/tex] where x is the length of the sides of the base.
Our goal is to find the absolute minimum of the the surface area function on the interval [tex](0,\infty)[/tex] and the dimensions of the box with minimum surface area.
1. To find the absolute minimum you must find the derivative of the surface area ([tex]S'(x)[/tex]) and find the critical points of the derivative ([tex]S'(x)=0[/tex]).
[tex]\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}[/tex]
Next,
[tex]2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120[/tex]
There is a undefined solution [tex]x=0[/tex] and a real solution [tex]x=2\sqrt[3]{15}[/tex]. These point divide the number line into two intervals [tex](0,2\sqrt[3]{15})[/tex] and [tex](2\sqrt[3]{15}, \infty)[/tex]
Evaluate S'(x) at each interval to see if it's positive or negative on that interval.
[tex]\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}[/tex]
An extremum point would be a point where f(x) is defined and f'(x) changes signs.
We can see from the table that f(x) decreases before [tex]x=2\sqrt[3]{15}[/tex], increases after it, and is defined at [tex]x=2\sqrt[3]{15}[/tex]. So f(x) has a relative minimum point at [tex]x=2\sqrt[3]{15}[/tex].
To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for [tex]x=2\sqrt[3]{15}[/tex].
[tex]\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}[/tex]
and for [tex]x=2\sqrt[3]{15}[/tex] we get:
[tex]2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0[/tex]
Therefore S(x) has a minimum at [tex]x=2\sqrt[3]{15}[/tex] which is:
[tex]S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2[/tex]
2. To find the third dimension of the box with minimum surface area:
We know that the volume is 60 [tex]ft^3[/tex] and the volume of a box with a square base is [tex]V=x^2h[/tex], we solve for h
[tex]h=\frac{V}{x^2}[/tex]
Substituting V = 60 [tex]ft^3[/tex] and [tex]x=2\sqrt[3]{15}[/tex]
[tex]h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft[/tex]
The dimension are the base edge [tex]x=2\sqrt[3]{15}\:ft[/tex] and the height [tex]h=\sqrt[3]{15} \:ft[/tex]
