The dimensions of the box are 10 ft and 5 ft
The maximum volume is 500 ft³
Step-by-step explanation:
A rectangle box with
Surface area of a box without top (SA) = perimeter of base × height + area of the base
Volume of a box (V) = base area × height
Assume that the length of the side of the square base is x and the height of the box is y
∵ It needs to be made using 300 ft² of material
∴ The surface area of the box is 300 ft²
∵ Its base is a square of side length x ft
∴ Perimeter of the base = 4 × x = 4 x
∴ Area of the base = x²
∵ The height of the box = y ft
∵ SA = perimeter of base × height + area of the base
∵ SA = (4x)(y) + x²
∴ SA = 4xy + x²
∵ SA of the box = 300 ft²
- Equate the two expressions of SA
∴ 4xy + x² = 300
Now let us find y in terms of x
- Subtract x² from both sides
∴ 4xy = 300 - x²
- Divide each term by 4x to find y
∴ [tex]y=\frac{75}{x}-\frac{1}{4}x[/tex]
∵ V = area of the base × height
∴ V = x² × y = x²y
- Substitute y by the equation of it above
∴ [tex]V=x^{2}(\frac{75}{x}-\frac{1}{4}x)[/tex]
∴ [tex]V=75x-\frac{1}{4}x^{3}[/tex]
∵ The volume of the box is greatest
- That means differentiate V and equate it by 0
∵ [tex]\frac{dV}{dx}=75-\frac{3}{4}x^{2}[/tex]
∵ [tex]\frac{dV}{dx}=0[/tex] ⇒ greatest volume
∴ [tex]75-\frac{3}{4}x^{2}=0[/tex]
- Subtract 75 from both sides
∴ [tex]-\frac{3}{4}x^{2}=-75[/tex]
- Divide both sides by [tex]-\frac{3}{4}[/tex]
∴ x² = 100
- Take √ for both sides
∴ x = 10
Substitute the value of x in the equation of y
∵ [tex]y=\frac{75}{10}-\frac{1}{4}(10)[/tex]
∴ y = 5
The dimensions of the box are 10 ft and 5 ft
∵ [tex]V=75x-\frac{1}{4}x^{3}[/tex]
∵ x = 10
∴ [tex]V=75(10)-\frac{1}{4}(10)^{3}[/tex]
∴ [tex]V=750-\frac{1}{4}(1000)[/tex]
∴ V = 750 - 250
∴ V = 500 ft³
The maximum volume is 500 ft³
Learn more:
You can learn more about the volume in brainly.com/question/6443737
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