Question

All boxes with a square​ base, an open​ top, and a volume of 220 ft cubed have a surface area given by ​S(x)equalsx squared plus StartFraction 880 Over x EndFraction ​, where x is the length of the sides of the base. Find the absolute minimum of the surface area function on the interval ​(0,infinity​). What are the dimensions of the box with minimum surface​ area?

Answer 1

Answer:

Length of the sides of the base (x) = 7.606 ft

Height (h) = 3.802 ft

The minimum surface area is 173.55 ft²

Step-by-step explanation:

Surface area is given by:

$S(x) = x^2+\frac{880}{x}$

The value of x for which the derivate of the surface area function is zero, is the length of the sides of the base that minimizes surface area:

$S(x) = x^2+\frac{880}{x} \\\frac{dS(x)}{dx}=0=2x-\frac{880}{x^2}\\x^3=440\\x=7.606\ ft$

The height of the box is given by:

$V=hx^2\\220 =h*7.606^2\\h=3.802\ ft$

The dimensions of the box with minimum surface area are:

Length of the sides of the base (x) = 7.606 ft

Height (h) = 3.802 ft

The absolute minimum is:

$S(x) = 7.606^2+\frac{880}{7.606}\\S_{min}=173.55\ ft^2$

The minimum surface area is 173.55 ft²

Answer 2

Answer:

The absolute minimum of the surface area$=173.55 ft^2$

At the minimum surface​ area,

• Base length=7.61 feet
• Height of 3.8 feet.

Step-by-step explanation:

Volume of the box =220 cubic feet.

$\text{Surface Area, } S(x)=x^2+\dfrac{880}{x}$

To find the absolute minimum of the surface area function on the interval ​$(0,\infty)$, we take the derivative of S(x) and solve for its critical points.

$S(x)=\dfrac{x^3+880}{x}\\S'(x)=\dfrac{2x^3-880}{x^2}\\ Setting the derivative equal to 0\\S'(x)=\dfrac{2x^3-880}{x^2}=0\\2x^3-880=0\\2x^3=880\\ Divide both sides by 2\\x^3=440$

Take the cube root of both sides

$x=\sqrt[3]{440}\\ x=7.61 ft$

Therefore, the absolute minimum of the surface area function on the interval $(0,\infty)$, is:

$S(x)=\dfrac{7.61^3+880}{7.61}\\\\=173.55 ft^2$

Since the volume of the box =220 cubic feet

$V=x^2h\\220=7.61^2 \times h\\h=220 \div 7.61^2\\h=3.80 ft$

The dimensions of the box with the minimum surface​ area are base length of 7.61 feet and height of 3.8 feet.