Answer:
Step-by-step explanation:
From the given information:
a)
Assuming the shape of the base is square,
suppose the base of each side = x
Then the perimeter of the base of the square = 4x
Suppose the length of the package from the base = y; &
the height is also = x
Now, the restriction formula can be computed as:
y + 4x ≤ 99
The objective function:
i.e maximize volume V = l × b × h
V = (y)*(x)*(x)
V = x²y
b) To write the volume as a function of x, V(x) by equating the derived formulas in (a):
y + 4x ≤ 99 --- (1)
V = x²y --- (2)
From equation (1),
y ≤ 99 - 4x
replace the value of y into (2)
V ≤ x² (99-4x)
V ≤ 99x² - 4x³
Maximum value V = 99x² - 4x³
At maxima or minima, the differential of [tex]\dfrac{d }{dx}(V)=0[/tex]
[tex]\dfrac{d}{dx}(99x^2-4x^3) =0[/tex]
⇒ 198x - 12x² = 0
[tex]12x \Big({\dfrac{33}{2}-x}}\Big)=0[/tex]
By solving for x:
x = 0 or x = [tex]\dfrac{33}{2}[/tex]
Again:
V = 99x² - 4x³
[tex]\dfrac{dV}{dx}= 198x -12x^2 \\ \\ \dfrac{d^2V}{dx^2}=198 -24x[/tex]
At x = [tex]\dfrac{33}{2}[/tex]
[tex]\dfrac{d^2V}{dx^2}\Big|_{x= \frac{33}{2}}=198 -24(\dfrac{33}{2})[/tex]
[tex]\implies 198 - 12 \times 33[/tex]
= -198
Thus, at maximum value;
[tex]\dfrac{d^2V}{dx^2}\le 0[/tex]
Recall y = 99 - 4x
when at maximum x = [tex]\dfrac{33}{2}[/tex]
[tex]y = 99 - 4(\dfrac{33}{2})[/tex]
y = 33
Finally; the volume V = x² y is;
[tex]V = (\dfrac{33}{2})^2 \times 33[/tex]
[tex]V =272.25 \times 33[/tex]
V = 8984.25 inches³
