A fish tank in the shape of a cuboid has length 320 cm. Its length is twice that of its width. to enhance viewing the area of the four vertical faces should be maximized. Find the optimum viewing area of a fish tank that is fixed to the wall so that the area of three faces only should be considered
To find the optimum viewing area of the fish tank, we need to maximize the area of the three faces that are visible when the tank is fixed to the wall.
Let's start by assigning variables. Let's say the width of the tank is x cm. Given that the length is twice the width, the length would be 2x cm.
The formula to find the surface area of a cuboid is: Surface Area = 2lw + 2lh + 2wh.
Since we want to maximize the area of three faces only, we can ignore one face. In this case, we'll ignore the bottom face, as it is not visible when the tank is fixed to the wall. Therefore, the formula becomes: Surface Area = 2lw + 2lh.
Now, let's substitute the values into the equation:
Surface Area = 2(2x)(x) + 2(2x)(x)
= 4x^2 + 4x^2
= 8x^2
To find the optimum viewing area, we need to find the maximum value of this equation.
To do that, we can take the derivative of the equation with respect to x and set it equal to zero.
d(Surface Area)/dx = 16x
Setting the derivative equal to zero:
16x = 0
Solving for x:
x = 0
Since x cannot be zero (as it represents the width of the tank), we can conclude that there is no maximum value for the viewing area when considering three faces only.
Therefore, we cannot find an optimum viewing area for this fish tank that is fixed to the wall while considering three faces only.