A line has an x-int. of 8 and a y-int. of 12. Determine the dimensions of the largest rectangle with two sides on the x and y axes that is tangent to the line.
2006-11-06 15:41:58 · 3 answers · asked by C 1 in Science & Mathematics ➔ Mathematics
First, you need to determine the equation of that line. You can see that the slope is -12/8 = -1.5, and you already know the y-intercept is 12, so the equation is y = -1.5x + 12.
Now you need to pick the tangency point, and calculate the area of the rectangle. If you choose an arbitrary value for x, then the side on the x-axis has length x, and the side on the y-axis has length y = -1.5x + 12. The product, therefore, is x(-1.5x + 12) = -1.5x^2 + 12x.
You need to maximize the area. This requires differentiation. The min or max value occurs when the derivative is zero. In this case, it will be a max, because the coefficient of x^2 is negative. The derivative is -3x + 12. Setting it equal to zero, -3x + 12 = 0 ==> x = 4. When x = 4, y = -1.5(4) + 12 = 6. So the dimensions are 4 x 6, with 4 on the x-axis and 6 on the y-axis.
2006-11-06 15:49:15 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋
the linethat crosses x = 8 and y = 12 is y=-(3/2)x+12 and the largest rectangle would be 6 on the y axis by 4 on the axis axis from the origin (0,0) being tangent at (4,6).
2006-11-06 15:50:36 · answer #2 · answered by bourqueno77 4 · 0⤊ 0⤋
Your textbook probably has a page or two at the beginning of the section explaing the methods for solving equations like this. I hope u dont end up on here some day asking "Should I run for president like everyone says?"
2006-11-06 15:44:25 · answer #3 · answered by James M 2 · 0⤊ 0⤋