Answer:
360
Step-by-step explanation:
1/30 the explanation is joe
To solve this problem you must follow the proccedure below:
a<span>. Find the perimeter and area of the cracker remaining:
The perimeter of a quarter circle is:
P=(</span>πr/2)+2r
P=(πx3 cm/2)+2(3 cm)
P=10.71 cm
The perimeter of <span>the cracker remaining is:
</span><span>
Pt=3 cm+6 cm+3 cm+10.71 cm
Pt=22.71 cm
The area of a quarter circle is:
A=</span>πr²/4
A=π(3 cm)²/4
A=7.06 cm²
<span>
The area of</span><span>of the cracker remaining is:
</span><span>
At=Area of a square-Area of quarter circle
At=L</span>²-(πr²/4)
At=(6 cm)²-(π(3 cm)²/4)
At=28.93 cm²
<span>
b. About how many bites can you get from the entire cracker?</span>
Number of bites=L²/(π(3 cm)²/4)
Number of bites=5
Answer:
![h = \sqrt[3]{\frac{49V}{4}}](https://tex.z-dn.net/?f=h%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B49V%7D%7B4%7D%7D)
Step-by-step explanation:
Represent the volume of the box with V and the dimensions with l, b and h.
The volume (V) is:

Make h the subject of the formula

The surface area (S) of the aquarium is:

Where lb represents the area of the base (i.e. slate):
The cost (C) of the surface area is:



Substitute
for h in the above equation



Differentiate with respect to l and with respect to b


To solve for b and l, we equate both equations and set l to b (to minimize the cost)


By comparison:

becomes

Cross Multiply

Solve for l

![l = \sqrt[3]{\frac{2V}{7}}](https://tex.z-dn.net/?f=l%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B2V%7D%7B7%7D%7D)
Recall that: 
![b = \sqrt[3]{\frac{2V}{7}}](https://tex.z-dn.net/?f=b%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B2V%7D%7B7%7D%7D)
Also recall that:

![h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7BV%7D%7B%5Csqrt%5B3%5D%7B%5Cfrac%7B2V%7D%7B7%7D%7D%2A%5Csqrt%5B3%5D%7B%5Cfrac%7B2V%7D%7B7%7D%7D%7D)
![h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7BV%7D%7B%5Csqrt%5B3%5D%7B%5Cfrac%7B4V%5E2%7D%7B49%7D%7D%7D)
Apply law of indices
![h = \sqrt[3]{\frac{49V^3}{4V^2}}](https://tex.z-dn.net/?f=h%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B49V%5E3%7D%7B4V%5E2%7D%7D)
![h = \sqrt[3]{\frac{49V}{4}}](https://tex.z-dn.net/?f=h%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B49V%7D%7B4%7D%7D)
The dimension that minimizes the cost of material of the aquarium is:
![h = \sqrt[3]{\frac{49V}{4}}](https://tex.z-dn.net/?f=h%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B49V%7D%7B4%7D%7D)