A closed cardboard box is designed to hold a volume of 3 288 cm . The length is 3 cm and the width x cm .Show that the total surface area 2 A cm is given by x A x 768 6 2 and find the dimensions of the box which will make A a minimum.
A closed cardboard box is designed to hold a volume of 288 cm^3 . The length is 3 cm and the width x cm .Show that the total surface area A cm^2 is given by A=6x^2+768/x and find the dimensions of the box which will make A a minimum. [8 Marks]
I have no idea what "x A x 768 6 2" means, or what the two missing characters are, but if the height is h, then
3xh = 3288
A = 2(3x + (3+x)h) = 2(3x+(3+x) * 3288/(3x)) = 6x + 2192 + 6576/x
dA/dx = 6 - 6576/x^2 = 6(x^2-1096)/x^2
dA/dx=0 when x=2√274 and h = 2√274
max volume (min area) occurs when the box is as close as possible to a cube. Since we were given 3 as one dimension, then the other dimensions turn out to be a square.
A closed cardboard box is designed to hold a volume of 3 288 cm . The length is 3 cm and the width x cm .Show that the total surface area 2 A cm is given by x A x 768 6 2 and find the dimensions of the box which will make A a minimum.
To solve this problem, let's break it down step by step.
1. Let's start by finding the height of the box. We know that the volume of the box is given by the formula V = lwh, where V is the volume, l is the length, w is the width, and h is the height. In this case, V = 3,288 cm³ and l = 3 cm. We can rearrange the formula to solve for h: h = V / (lw).
Substituting the given values, we have h = 3,288 cm³ / (3 cm * w).
2. Now, let's find the surface area of the box. The total surface area of a rectangular box is given by the formula A = 2lw + 2lh + 2wh. Again, substituting the known values, we have A = 2(3 cm * w) + 2(3 cm * h) + 2(w * h).
Simplifying, we have A = 6w + 6h + 2wh.
3. Next, let's substitute the expression we found for h in step 1 into the equation for A:
A = 6w + 6(3,288 cm³ / (3 cm * w)) + 2(w * (3,288 cm³ / (3 cm * w))).
Simplifying further, A = 6w + 6(1,096 cm² / w) + 2(1,096 cm²).
4. Now, let's simplify the equation:
A = 6w + 6,576 cm² / w + 2,192 cm².
5. To find the dimensions of the box that will minimize A, we need to find the value of w that minimizes the equation. We can do this by taking the derivative of A with respect to w, setting it equal to zero, and solving for w.
dA/dw = 6 - 6,576 cm² / w².
Setting dA/dw = 0, we get 6 - 6,576 cm² / w² = 0.
Simplifying further, we have 6w² - 6,576 cm² = 0.
Dividing by 6, we get w² - 1,096 cm² = 0.
Solving for w, we find w = √1,096 cm.
6. Finally, substituting this value of w back into the equation for A:
A = 6w + 6(1,096 cm² / w) + 2(1,096 cm²).
A = 6√1,096 cm + 6(1,096 cm / √1,096 cm) + 2(1,096 cm).
Simplifying, we get A = 12√1,096 cm + 3,288 cm.
7. To express A in the given form (xA + 7686), we can rewrite A in terms of x:
A = 12√1,096 cm + 3,288 cm = 12x + 7686.
Comparing coefficients, we can conclude that x = √1,096 cm.
Therefore, the dimensions of the box that will minimize A are length = 3 cm, width = √1,096 cm, and height = 3,288 cm / (3 cm * √1,096 cm).