Answer:
x=2.125
y=0
C=19.125
Step-by-step explanation:
To solve this problem we can use a graphical method, we start first noticing the restrictions 
and 
, which restricts the solution to be in the positive quadrant. Then we plot the first restriction 
shown in purple, then we can plot the second one 
shown in the second plot in green.
The intersection of all three restrictions is plotted in white on the third plot. The intersection points are also marked.
So restrictions intersect on (0,0), (0,1.7) and (2.215,0). Replacing these coordinates on the objective function we get C=0, C=11.9, and C=19.125 respectively. So The function is maximized at (2.215,0) with C=19.125.
Answer:
4times tall
Step-by-step explanation:
Volume of the boxes = Base area × height
Volume of the first box V1 = A1h1
Given the base of the first box to be 5cm, the base area:
A1 = 5cm×5cm = 25cm²
Volume of the first box V1 = 25h1... 1
Similarly, volume of the second box
V2 = A2h2
Given the base of the second box to be 10cm, the base area:
A2= 10cm×10cm = 100cm²
Volume of the second box
V2 = 100h2... 2
If the two boxes have the same volume, then V1 = V2
25h1 = 100h2
divide both sides by 25
25h1/25 = 100h2/25
h1 = 4h2
Since the height of the smaller box is represented as h1, then the height of the smaller base is 4 times tall.
Step 1: x/3-8=2
Step 2: x/-5=2
Step 3: -10/-5=2
Step 4: x=-10
Step 1: First, subtract 8 from 3. 3-8=-5.
Step 2: Replace 3-8 with -5.
Step 3: To find x, multiply -5 and 2. The answer is -10.
Step 4: Write out the final answer, which is x=-10
now, we get critical points from zeroing out the derivative, and also from zeroing out the denominator, but those at the denominator are critical points where the function is not differentiable, namely a sharp spike or cusp or an asymptote.
so, from zeroing out the derivative we get no critical points there, from the denominator we get x = 8, but can't use it because f(x) is undefined.
therefore, we settle for the endpoints, 4 and 6,
f(4) =3 and f(6) = 7
doing a first-derivative test, we see the slope just goes up at both points and in between, but the highest is f(6), so the absolute maximum is there, while we can take say f(4) as the only minimum and therefore the absolute minumum as well.
