Answer: [tex]V_{(max)}=6744.60 cm^{3}[/tex]
Step-by-step explanation:
Here we have to find the value for [tex]x[/tex] in order to have the maximum volume ([tex]V_{max}[/tex]) of the box, where the area of the box is given by:
[tex]A_{(x)}=1700=x^{2} + 4hx[/tex] (1)
And the volume:
[tex]V_{(x)}=(x)(x)(h)=x^{2}h[/tex] (2)
Isolating [tex]h[/tex] from (1):
[tex]h=\frac{1700-x^{2}}{4x}[/tex] (3)
Substituting (3) in (2):
[tex]V_{(x)}=(x)(x)(h)=x^{2}(\frac{1700-x^{2}}{4x})=-\frac{1}{4}x^{3}+425 x[/tex] (4)
So, we have to derive this expression, then make it equal to zero and solve the equation to find the critical point:
[tex]V_{x}^{'}=-\frac{3}{4}x^{2} +425[/tex] (5)
Making [tex]V_{h}^{'}=0[/tex]:
[tex]-\frac{3}{4}x^{2} +425=0[/tex] (6)
The critical point is:
[tex]x=23.80 cm[/tex] (7)
This means the maximum Volume is when [tex]x=23.80 cm[/tex]. Now with this value we can find [tex]h[/tex]:
[tex]h=\frac{1700 cm^{2}-(23.80 cm)^{2}}{4(23.80 cm)}=11.907 cm[/tex] (8)
Hence:
[tex]V_{(max)}=(23.8 cm)^{2}(11.90 cm)=6744.60 cm^{3}[/tex] (9) This is the maximum volume of the box
