Answer :
Let's solve the question step-by-step.
10.1.1 Show that the surface area of the block is given by [tex]A = 2x^2 + \frac{32}{x}[/tex]
Understanding the Shape: A square right prism has a square base. Suppose each side of the base is [tex]x[/tex] meters, and the height of the prism is [tex]h[/tex] meters. The formula for the volume of the prism is:
[tex]V = x^2 \, h[/tex]
Given that the volume is 8 m³:
[tex]x^2 \, h = 8[/tex]
From this, we can express [tex]h[/tex] as:
[tex]h = \frac{8}{x^2}[/tex]Surface Area: The surface area [tex]A[/tex] consists of:
- The area of the top and bottom, which are both [tex]x^2[/tex] (so [tex]2x^2[/tex]).
- The area of the four sides which is [tex]4x \cdot h[/tex].
Substituting [tex]h = \frac{8}{x^2}[/tex] into the equation for the lateral surface area:
[tex]4x \cdot h = 4x \cdot \frac{8}{x^2} = \frac{32}{x}[/tex]So, the total surface area [tex]A[/tex] is:
[tex]A = 2x^2 + \frac{32}{x}[/tex]
10.1.2 Calculate the dimensions of the block that will ensure that a minimum quantity of paint will be used.
To minimize the surface area, we need to take the derivative of [tex]A[/tex] with respect to [tex]x[/tex], set it to zero, and solve for [tex]x[/tex].
Derivative of the Surface Area
[tex]A = 2x^2 + \frac{32}{x}[/tex]The derivative, [tex]\frac{dA}{dx}[/tex], is:
[tex]\frac{dA}{dx} = 4x - \frac{32}{x^2}[/tex]Find Critical Points
Set [tex]\frac{dA}{dx} = 0[/tex]:
[tex]4x - \frac{32}{x^2} = 0[/tex]
[tex]4x = \frac{32}{x^2}[/tex]
[tex]4x^3 = 32[/tex]
[tex]x^3 = 8[/tex]
[tex]x = 2[/tex]Find [tex]h[/tex] from [tex]x[/tex]
Substituting [tex]x = 2[/tex] back into [tex]h = \frac{8}{x^2}[/tex]:
[tex]h = \frac{8}{2^2} = 2[/tex]
Thus, the dimensions of the block that minimize the surface area, and hence the paint used, are 2 meters by 2 meters by 2 meters (a cube).
