Respuesta :
Answer:
Step-by-step explanation:
Refer the attached picture for complete question
a) V(x)=x(11-2x)(17-2x)
B(x)=140
Side and volume cannot be negative
So, x> 0
11-2x>0 and 17-2x>0
[tex]x<\frac{11}{2}[/tex] and [tex]x< \frac{17}{2}[/tex]
Domain for V(x) :[tex]0<x<\frac{11}{2}[/tex]
b)For finding the greatest Volume
[tex]\frac{dV}{dx}=0[/tex]
[tex]V(x)=x(11-2x)(17-2x)=4x^3-56x^2+187x[/tex]
[tex]\frac{dV}{dx}=12x^2-112x+187=0\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{112\pm\sqrt{(112)^2-4(12)(187)}}{2(12)}[/tex]
x=7.155 , 2.177
Consider the graph
So, Volume at 7.155 is negative
Thus x≈2 gives the greatest volume
c)
Consider the graph .
It is increasing from [tex]-1 \leq x \leq 2[/tex] and [tex]x \geq 7[/tex]
So, Graph increasing interval is [-1,2] ∪[7,∞)
d)
At the point of intersection, both graph has the same value .
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