Respuesta :
Answer:
5040.58 cubic cm.
Step-by-step explanation:
Let x represent length and width of the square base and y represent height of box.
We have been that 1400 square centimeters of material is available to make a box with a square base and an open top. We are asked to find the largest possible volume of the box.
The total surface area of box will be equal to area of square base plus area of 4 sides.
[tex]\text{Total surface area of box}=x^2+4xy[/tex]
[tex]1400=x^2+4xy[/tex]
Let us solve for y.
[tex]1400-x^2=4xy[/tex]
[tex]\frac{1400-x^2}{4x}=y[/tex]
Volume of the box would be equal to area of base times height.
[tex]V=x^2y[/tex]
Upon substituting [tex]\frac{1400-x^2}{4x}=y[/tex] in volume equation, we will get:
[tex]V=x^2(\frac{1400-x^2}{4x})\\\\ V=\frac{1400x^2-x^4}{4x}\\\\V=\frac{1400x-x^3}{4}[/tex]
We need to maximize the volume, so need to find critical points of derivative of volume function.
[tex]V'=\frac{1}{4}(\frac{d}{dx}(1400x)-\frac{d}{dx}x^3))[/tex]
Using power rule, we will get:
[tex]V'=\frac{1}{4}(1400-3x^2)[/tex]
[tex]V'=\frac{1400-3x^2}{4}[/tex]
TO find critical points, we need to equate derivative with 0.
[tex]\frac{1400-3x^2}{4}=0[/tex]
[tex]1400-3x^2=0\\\\-3x^2=-1400\\\\x^2=\frac{1400}{3}\\\\x^2=466.666666\\\\x=\pm\sqrt{466.666666}\\\\x=\pm 21.60246[/tex]
Since length cannot be negative, therefore, length would be 21.60 cm.
To find maximum volume, we need to substitute [tex]x=21.60246[/tex] in volume function.
[tex]V=\frac{1400(21.60246)-(21.60246)^3}{4}[/tex]
[tex]V=\frac{30243.444-10081.139604958566936}{4}[/tex]
[tex]V=\frac{20162.304395041433064}{4}[/tex]
[tex]V=5040.576098760358266\approx 5040.58[/tex]
Therefore, the maximum volume of the box would be 5040.58 cubic cm.
The largest possible volume of a box is "5040.58 cm³".
According to the question,
The surface area,
- 1400 cm²
Let,
- Length be "x".
- Width be "x".
- Height be "y".
then,
→ [tex]x^2+4xy = 1400[/tex]
[tex]y = \frac{1400-x^2}{4x}[/tex]
The volume of box:
→ [tex]V = x^2y[/tex]
[tex]= x^2(\frac{1400-x^2}{4x} )[/tex]
[tex]= 350x-\frac{x^3}{4}[/tex]
then,
→ [tex]\frac{dV}{dx} = 350-\frac{3x^2}{4}[/tex]
For maxima and minima, [tex]\frac{dV}{dx} =0[/tex]
→ [tex]350-\frac{3x^2}{4} =0[/tex]
[tex]3x^2=1400[/tex]
[tex]x = \frac{10\sqrt{42} }{3}[/tex]
Therefore,
The maximum volume (V) will be:
= [tex]350x-\frac{x^3}{4}[/tex]
= [tex]350(\frac{10\sqrt{42} }{3} )-\frac{1}{4}(\frac{10\sqrt{42} }{3} )[/tex]
= [tex]5040.58 \ cm^3[/tex]
Thus the above approach is right.
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