Answer:
1) [tex]S = 2\cdot w\cdot l - 8\cdot x^{2}[/tex], 2) The domain of S is [tex]0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}[/tex]. The range of S is [tex]0 \leq S \leq 2\cdot w \cdot l[/tex], 3) [tex]S = 176\,in^{2}[/tex], 4) [tex]x \approx 4.528\,in[/tex], 5) [tex]S = 164.830\,in^{2}[/tex]
Step-by-step explanation:
1) The function of the box is:
[tex]S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)[/tex]
[tex]S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w[/tex]
[tex]S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x[/tex]
[tex]S = 2\cdot w\cdot l - 8\cdot x^{2}[/tex]
2) The maximum cutout is:
[tex]2\cdot w \cdot l - 8\cdot x^{2} = 0[/tex]
[tex]w\cdot l - 4\cdot x^{2} = 0[/tex]
[tex]4\cdot x^{2} = w\cdot l[/tex]
[tex]x = \frac{\sqrt{w\cdot l}}{2}[/tex]
The domain of S is [tex]0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}[/tex]. The range of S is [tex]0 \leq S \leq 2\cdot w \cdot l[/tex]
3) The surface area when a 1'' x 1'' square is cut out is:
[tex]S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}[/tex]
[tex]S = 176\,in^{2}[/tex]
4) The size is found by solving the following second-order polynomial:
[tex]20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}[/tex]
[tex]20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}[/tex]
[tex]8\cdot x^{2} - 164\,in^{2} = 0[/tex]
[tex]x \approx 4.528\,in[/tex]
5) The equation of the box volume is:
[tex]V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x[/tex]
[tex]V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x[/tex]
[tex]V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}[/tex]
[tex]V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}[/tex]
[tex]V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}[/tex]
The first derivative of the function is:
[tex]V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}[/tex]
The critical points are determined by equalizing the derivative to zero:
[tex]12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0[/tex]
[tex]x_{1} \approx 4.952\,in[/tex]
[tex]x_{2}\approx 1.548\,in[/tex]
The second derivative is found afterwards:
[tex]V'' = 24\cdot x - 78\,in[/tex]
After evaluating each critical point, it follows that [tex]x_{1}[/tex] is an absolute minimum and [tex]x_{2}[/tex] is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:
[tex]x \approx 1.548\,in[/tex]
The surface area of the box is:
[tex]S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}[/tex]
[tex]S = 164.830\,in^{2}[/tex]
Answer:
A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constructed by cutting four equal-sized surface corners from an 8-inch by 11.5 inch sheet of cardboard and folding up the sides.
1. Determine a function that relates the total surface area, s, (measured in square inches) of the open box to the size of the square cutout x (measured in inches).
2. What is the domain and range of the function s?
3. What is the surface area when a 1" x 1" square is cut out?
4. What size square cutout will result in a surface area of 20 in?
5. What is the surface area of the box when the volume is maximized? (Calculator)
Step-by-step explanation:
