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Kagome_gurl8
Miguel is designing shipping boxes that are rectangular prisms. One shape of the box with height h in feet, has a volume defined by the function V(h) = h(h - 8)(h - 6). Graph the function. What is the maximum volume for the domain 0 < h < 8? Round to the nearest cubic foot.
I tried to graph it and everytime it says "error"
Use x as the variable, not h
\[V(h) = h(h-8)(h-6)\]\[V(h) = (h^2-8h)(h-6)\]\[V(h) = h^3-6h^2-8h^2+48h\]\[V(h) = h^3-14^2+48h\] What you have here is a 3rd degree polynomial. To find the maximum volume on a closed set [0,8] we first test the endpoints to see what values of V we obtain. So for h = 0 (its pretty obvious Volume =0) \[V(0)=0^3-14(0)^2+48(0) = 0 \] for h=8, we have \[V(8)=8^3-14(8)^2+48(8) = 0 \]\[V(8)=512-896+384\]\[V(8)= 0 \] So we have that the extrema created by the set boundary do not give maximum values. To test the function in between the domain set, we take the first derivative of the function:\[\frac{ d }{ dh } V(h)=3h^2−28h+48\] and solve for when this derivative is 0, which would indicate the presence of a maximum or minimum. \[0 = 3h^2−28h+48\] using the quadratic formula, we find the roots are h=7.07 and h=2.26. At these height values we either have a maximum or minimum, and since they both are within the domain, they are up for consideration. Testing each point (in the original equation) we have that h=7.07 gives us a volume of -7.04. This is clearly nonsense, since volume cannot be negative. We then test 2.26, which gives us a volume of 48.5. Since we have tested for all possible extrema and since this is the largest value of V on the interval [0,8], we have that the maximum volume, as determined by the equation, will be 48.5 when the height of the rectangular prism is 2.26
This can all be clearly seen in the graph of the function posted by Hero. A minimum exists at 7 and a maximum at 2.26, and at the endpoints we have that V=0
Woah.... Amazing thank you so much!